项目经理证怎么考取,企业网站怎样做seo优化 应该如何做,可做实名认证的网站,烟台网站制作设计题目描述 对不同的迷宫进行算法问题#xff0c;广度优先、深度优先、以及人工智能上介绍的一些算法#xff1a;例如A*算法#xff0c;蚁群算法等。 基本要求#xff1a; #xff08;1#xff09;从文件读入9*9的迷宫#xff0c;设置入口和出口#xff0c;分别采用以上方…题目描述 对不同的迷宫进行算法问题广度优先、深度优先、以及人工智能上介绍的一些算法例如A*算法蚁群算法等。 基本要求 1从文件读入9*9的迷宫设置入口和出口分别采用以上方法输出从入口到出口的一条路径 2从文件读入9*9的迷宫设置入口和出口试着求出所有可行路径使用一种算法求解即可。 3从文件读入9*9的迷宫设置入口和出口并设置通路中的代价试从上述结构和算法中选择合适的方式求出从入口到出口的最短路径。 4随机生成90*90的迷宫以及入口和出口并设置通路中的代价请设计选择至少两种方法求出从入口到出口代价最小的路径并对这两种方法进行算法时间效率和空间效率的比较从实验结果或数据说明结论。 5尝试随机生成更大的迷宫地图尽量大以及入口、出口以及通路中的代价请设计选择至少两种方法求出从入口到出口的代价最小的路径并对这两种方法进行算法时间效率和空间效率的比较从实验结果或数据说明结论。 高级要求 具有人机交互的图形界面 前言
迷宫问题是一个经典的人工智能问题通过模拟人们在迷宫中的寻路过程可以探索和比较不同的寻路算法的优劣。
迷宫问题的对比实验研究可以帮助我们了解不同寻路算法的原理和工作机制。在实验中我们可以设计不同的迷宫环境和障碍物布局然后应用不同的寻路算法来解决迷宫问题。通过对比实验结果我们可以分析不同算法在不同情况下的表现从而深入理解其原理和特点。
迷宫问题的对比实验研究可以为实际应用场景提供参考和指导。在现实生活中我们也经常会遇到需要寻找最优路径的问题比如机器人导航、物流配送等。通过对比实验研究我们可以评估不同算法在实际场景中的性能表现从而选择最适合的算法来解决实际问题。
迷宫问题的对比实验研究还可以为算法改进和优化提供思路和方向。通过对比不同算法的实验结果我们可以发现算法的优点和不足之处进而提出改进和优化的方法。这有助于推动寻路算法的发展提高其在实际应用中的性能和效率。
综上迷宫问题的对比实验研究对于理解不同寻路算法的原理和工作机制、指导实际应用场景的选择以及促进算法改进和优化具有重要意义。 第1题 广度优先搜索
算法思想
从一个起始节点开始一层一层地访问所有相邻节点直到无法继续为止。具体来说BFS从入口点开始将其标记为已访问并加入一个队列中。然后它取出队首的节点访问所有与该节点相邻且未被访问过的节点并将这些节点标记为已访问加入到队列中。这个过程会一直持续到队列为空即所有可达的节点都已访问过。
主要步骤如下
1. 从入口点开始将其标记为已访问
2. 将入口点加入一个队列中
3. 当队列不为空时执行以下操作 - 取出队首的节点 - 如果该节点是出口点那么找到了一条路径 - 否则将该节点的所有未被访问过的相邻节点标记为已访问并加入到队列中
4. 如果队列为空但还没有找到出口点那么说明没有路径可以通向出口点。 算法实现
// Created by Mr.Chen on 2023/12/24.
#include stdio.h
#include stdlib.h/*** 9*9迷宫(最外围是一堵墙)* 1代表墙0代表空格* 只有东西南北4个方向可以走*/
#define SIZE 9 //迷宫大小
int maze[SIZE][SIZE]; //迷宫
int path[SIZE][SIZE] {0}; //记录路径上的坐标标记值为1/*** 存储路径坐标的结点称为路径结点* 因为数量无法确定故使用链式结构*/
typedef struct Node
{int coordinate; //坐标struct Node* next;
}Node;typedef Node* PNode; //路径结点的指针类型typedef struct Queue
{PNode front; //队头指针指向队头元素PNode rear; //队尾指针指向队尾元素
}Queue;typedef Queue* LinkQueue; //链队列/*** 创建并初始化一个链队列*/
LinkQueue creatLinkQueue()
{LinkQueue linkQueue (LinkQueue)malloc(sizeof(Queue));linkQueue-frontNULL;linkQueue-rearNULL;return linkQueue;
}/*** 空队返回1非空返回0*/
int isEmpty(LinkQueue linkQueue)
{return linkQueue-front NULL;
}/*** 入队*/
void inQueue(LinkQueue linkQueue,int coordinate)
{PNode pNode (PNode)malloc(sizeof(Node));pNode-coordinate coordinate;pNode-next NULL;if(isEmpty(linkQueue)){linkQueue-front pNode;linkQueue-rear pNode;}else{//链队非空新的路径结点接在尾结点之后linkQueue-rear-next pNode;//更新尾指针linkQueue-rear pNode;}
}/*** 出队。返回队头元素的坐标并删除队头元素*/
int outQueue(LinkQueue linkQueue)
{int coordinate linkQueue-front-coordinate;PNode t linkQueue-front;linkQueue-frontlinkQueue-front-next; //改变指向跨越删除free(t);return coordinate;
}/*** 从文件中读取迷宫*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\1.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);
}/*** 打印迷宫和路径*/
void printfMaze()
{printf(迷宫以及路径\n■(墙) .(路) ♧(入口) ♤(出口) √(路径)\n);printf( );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{if(path[i][j] 1)printf(√ );elseprintf(. );}}printf(\n);}//打印路径坐标path[1][1] 1;path[SIZE - 2][SIZE - 2] 1;for(int i 0; i SIZE -1; i){for(int j 0; j SIZE - 1; j)if(path[i][j] 1)printf(-(%d,%d), i, j);}
}/*** 广度优先搜索一条从入口到出口的路径* param entryX 入口行坐标* param entryY 入口列坐标* param exitX 出口行坐标* param exitY 出口列坐标* param maze 迷宫* return 存在路径则返回1否则返回0*/
int mazeBSF(int entryX, int entryY, int exitX, int exitY)
{//迷宫的访问标记数组。未访问0 已访问1int mark[9][9] {0};//从入口开始探索故将入口标记为已访问mark[entryX][entryY] 1;//记录行走过程中前驱结点的坐标借助这些记录来回溯到起点int preX_record[SIZE][SIZE];int preY_record[SIZE][SIZE];//前驱结点的坐标初始化为-1for(int i 0; i SIZE; i){for(int j 0; j SIZE; j){preX_record[i][j]-1;preY_record[i][j]-1;}}//创建两个链队列//linkQueueX存储行坐标LinkQueue linkQueueX creatLinkQueue();//linkQueueX存储行坐标LinkQueue linkQueueY creatLinkQueue();//入口坐标入队inQueue(linkQueueX, entryX);inQueue(linkQueueY, entryY);mark[entryX][entryY] 1;//还有路走就一直循环while(!isEmpty(linkQueueX)){//队头元素出队获取根据地坐标。可理解为正在探访的位置的前驱坐标int preX outQueue(linkQueueX);int preY outQueue(linkQueueY);//每次都绕根据地一圈进行探索for(int mov 0; mov 4 ; mov){//4个移动方向按东、南、西、北的顺时针顺序依次探索下一个位置int direction[4][2]{{0,1},{1,0},{0,-1},{-1,0}};//正在探索的位置的坐标int X preX direction[mov][0];int Y preY direction[mov][1];//找到出口if (X exitX Y exitY){//记录出口的前驱位置preX_record[X][Y] preX;preY_record[X][Y] preY;//回溯while(X ! entryX Y ! entryY){preX preX_record[X][Y];preY preY_record[X][Y];path[preX][preY] 1; //记录路径X preX;Y preY;}return 1;}//此次探访到的坐标不是出口但它可以到达且尚未访问else if(maze[X][Y] 0 mark[X][Y] 0){//此次探访到的结点入队作为未来的根据地inQueue(linkQueueX,X);inQueue(linkQueueY,Y);//标记为已访问mark[X][Y]1;//记录此次探访结点的前驱preX_record[X][Y] preX;preY_record[X][Y] preY;}}}return 0;
}
int main()
{fileRead(); //从文件读取迷宫//将入口设置在左上角出口设置在右下脚。int entryX 1;int entryY 1;int exitX 7;int exitY 7;//搜索一条从入口到出口的路径int find mazeBSF(entryX,entryY,exitX,exitY);if(find 1)printfMaze(); //打印迷宫和路径elseprintf(找不到路径);return 0;
}
测试用例 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 输出结果 A*算法
算法思想
A*算法是一种启发式搜索算法它综合了广度优先搜索和贪心算法的思想以寻找最短路径为目标。这种算法在迷宫上表现良好能够高效地找到代价值最小路程最短的路径。
A*算法运用了一种评估函数值的方法来估算节点的代价该函数值是实际移动代价和预计代价的总和。预计代价是根据启发式函数计算的表示从当前节点到目标节点的预估距离。因此A*算法在选择下一个需要遍历的节点时会选取综合优先级最高的节点也就是代价值最小的节点。然而选择一个合适的启发式函数很重要不同的启发式函数可能导致不同的搜索结果。
主要步骤 1、创建一个优先队列来保存待处理的节点。每个节点都有一个f(n)值表示从起点经过当前节点到目标节点的预估最短路径长度。
2、将起点加入队列并将其f(n)值设为0。
3、重复以下步骤直到队列为空或找到出口 a.从队列中取出优先级最高的节点作为当前节点。 b.如果当前节点是出口则搜索结束。 c.否则计算当前节点的邻居节点并更新它们的f(n)值和路径长度。如果某个邻居节点的新路径长度比之前的更短则将其加入队列。
4、如果队列为空而没有找到目标节点则表示无法到达出口。 算法实现
队列操作、寻找邻居结点等其实可以利用广度优先搜索的函数实现笔者还没来得及整合。这里有许多优化空间读者可自行优化
// Created by Mr.Chen on 2023/12/29.
#include stdio.h
#include stdlib.h
#include stdbool.h#define SIZE 9 //迷宫大小
int maze[SIZE][SIZE];
bool visited[SIZE][SIZE] {false}; //访问标记数组
int path[SIZE][SIZE] {0}; //存储路径的数组// 定义迷宫的起点和终点
typedef struct {int x;int y;
} coordinate;coordinate start {1, 1};
coordinate end {SIZE-2, SIZE-2};// 定义优先级队列的节点
typedef struct {int f; // f(n) g(n) h(n)int g; // 从起点到当前节点的路径长度coordinate point; // 当前节点的坐标
} Node;// 比较两个节点的优先级
int compare(const void *a, const void *b) {return ((Node*)a)-f - ((Node*)b)-f;
}// 获取一个current节点的所有邻居节点
void getNeighbors(coordinate current, coordinate neighbors[]) {int idx 0;if (current.x 1) {neighbors[idx].x current.x - 1;neighbors[idx].y current.y;idx;}if (current.x SIZE - 2) {neighbors[idx].x current.x 1;neighbors[idx].y current.y;idx;}if (current.y 1) {neighbors[idx].x current.x;neighbors[idx].y current.y - 1;idx;}if (current.y SIZE - 2) {neighbors[idx].x current.x;neighbors[idx].y current.y 1;}
}// 计算启发函数曼哈顿距离
int heuristic(coordinate node, coordinate goal) {return abs(node.x - goal.x) abs(node.y - goal.y);
}// 使用A*算法求解最短路径
int Astar() {//初始化存储结点的队列Node queue[SIZE * SIZE];int front 0, rear 0;//初始化各结点到起点的距离表默认为无穷大int distances[SIZE][SIZE];for (int i 0; i SIZE; i) {for (int j 0; j SIZE; j) {distances[i][j] INT_MAX;}}distances[start.x][start.y] 0;//记录行走过程中前驱结点的坐标借助这些记录来回溯到起点int preX_record[SIZE][SIZE];int preY_record[SIZE][SIZE];//前驱结点的坐标初始化为-1for(int i 0; i SIZE; i){for(int j 0; j SIZE; j){preX_record[i][j]-1;preY_record[i][j]-1;}}// 创建起点节点计算启发函数并加入队列Node startNode {heuristic(start, end), 0, start};queue[rear] startNode;// 开始执行A*算法while (front ! rear) {// 从队列中取出优先级最高的节点qsort(queue front, rear - front, sizeof(Node), compare);Node current queue[front];// 如果当前节点是终点返回最短路径长度if (current.point.x end.x current.point.y end.y){int ansDistance distances[current.point.x][current.point.y];//回溯while(current.point.x ! start.x current.point.y ! start.y){int preX preX_record[current.point.x][current.point.y];int preY preY_record[current.point.x][current.point.y];path[preX][preY] 1; //记录路径current.point.x preX;current.point.y preY;}return ansDistance;}//获取当前节点的所有邻居节点coordinate neighbors[4];getNeighbors(current.point, neighbors);for (int i 0; i 4; i) {coordinate neighbor neighbors[i];if (maze[current.point.x][current.point.y] 0 visited[neighbor.x][neighbor.y] false) {visited[neighbor.x][neighbor.y] true;// 计算新的路径长度int newDist current.g 1;// 如果新的路径更短则更新距离表并加入队列if (newDist distances[neighbor.x][neighbor.y]) {distances[neighbor.x][neighbor.y] newDist;Node newNode {newDist heuristic(neighbor, end), newDist, neighbor};queue[rear] newNode;//记录此次探访结点的前驱preX_record[neighbor.x][neighbor.y] current.point.x;preY_record[neighbor.x][neighbor.y] current.point.y;}}}}// 没有找到最短路径返回-1表示失败return -1;
}/*** 从文件中读取迷宫*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\1.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);
}/*** 打印迷宫和路径*/
void printfMaze()
{printf(迷宫■(墙) .(路) ♧(入口) ♤(出口) √(路径)\n);printf( );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{if(path[i][j] 1)printf(√ );elseprintf(. );}}printf(\n);}//打印路径坐标path[1][1] 1;path[SIZE - 2][SIZE - 2] 1;for(int i 0; i SIZE -1; i){for(int j 0; j SIZE - 1; j)if(path[i][j] 1)printf(-(%d,%d), i, j);}printf(\n);
}int main()
{fileRead(); //从文件读取迷宫int shortestPathLength Astar();printfMaze();printf(最短路径的距离: %d\n, shortestPathLength);return 0;
} 测试用例 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 输出结果 深度优先搜索
算法思想
基本思想是从一个起始点开始沿着一条路径尽可能深地探索迷宫中的格子直到达到不能继续前进的格子然后回溯到前一个格子继续探索其他路径
具体来说在迷宫问题中运用深度优先搜索算法的主要步骤如下
1.从入口点开始将其标记为已访问
2.选择一个尚未访问的相邻格子作为新的起始点
3.重复第二步直到走到死胡同即路径最深处没有相邻的未访问的格子
4.如果还存在未访问的格子则回溯到上一个格子
5.重复第三步和第四步直到找到出口点或者所有可走的路径都已探寻完毕。 算法实现
// Created by Mr.Chen on 2023/12/23.#includestdio.h
#includestdlib.h/*** 9*9迷宫(最外围是一堵墙)* 1代表墙0代表空格* 只有东西南北4个方向可以走*/
#define SIZE 9 //迷宫大小
int maze[SIZE][SIZE]; //迷宫
int path[SIZE][SIZE] {0}; //记录路径上的坐标标记值为1/*** 存储路径坐标的结点称为路径结点* 因为数量无法确定故使用链式结构*/
typedef struct Node
{int coordinate; //坐标struct Node* next;
}Node;typedef Node* PNode; //路径结点的指针类型
typedef Node* LinkStack; //链栈指针,可代表栈顶指针/*** 创建一个链栈并返回栈顶指针*/
LinkStack creatLinkStack()
{LinkStack top (LinkStack)malloc(sizeof(Node));if(top ! NULL)top-next NULL;elseprintf(链栈创建失败);return top;
}/*** param top 坐标要入的栈的栈顶指针* param coordinate 入栈坐标*/
void push(LinkStack top, int coordinate)
{PNode pNode (PNode)malloc(sizeof(Node));if(pNode NULL)printf(入栈失败);else{pNode-coordinate coordinate;//头插法pNode-next top-next;top-next pNode;}
}/*** 空栈返回1非空返回0*/
int isEmpty(LinkStack top)
{return top-nextNULL ? 1 : 0;
}/*** 弹出栈顶元素的坐标*/
int pop(LinkStack top)
{int coordinate;coordinate top-next-coordinate; //获取栈顶元素的坐标//删除栈顶元素更新栈顶指针PNode t top-next;top-next t-next;free(t);return coordinate;
}/*** 从文件中读取迷宫*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文*/fp fopen(D:\\Projects\\ClionProjects\\mazeText\\1.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);
}/*** 打印迷宫*/
void printfMaze()
{printf(迷宫■(墙) .(路) ♧(入口) ♤(出口)\n);printf( );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{if(path[i][j] 1)printf(√ );elseprintf(. );}}printf(\n);}/*//打印路径坐标path[1][1] 1;path[SIZE - 2][SIZE - 2] 1;printf(\n);for(int i 0; i SIZE -1; i){for(int j 0; j SIZE - 1; j)if(path[i][j] 1)printf(-(%d,%d), i, j);}*/
}/*** 深度优先搜索一条从入口到出口的路径* param entryX 入口行坐标* param entryY 入口列坐标* param exitX 出口行坐标* param exitY 出口列坐标* param maze 迷宫* return 存在路径则返回1否则返回0*/
int mazeDFS(int entryX, int entryY, int exitX, int exitY)
{//迷宫的访问标记数组。未访问0 已访问1int mark[9][9] {0};//从入口开始探索故将入口标记为已访问mark[entryX][entryY] 1;//LinkStackX储存行坐标LinkStack LinkStack_X creatLinkStack();//LinkStackY储存列坐标LinkStack LinkStack_Y creatLinkStack();//将入口坐标入栈push(LinkStack_X,entryX);push(LinkStack_Y,entryY);while(!isEmpty(LinkStack_X)){int curX,curY; //当前位置的行坐标和列坐标curX pop(LinkStack_X);curY pop(LinkStack_Y);//4个移动方向按东、南、西、北的顺时针顺序依次探索下一个位置int direction[4][2]{{0,1},{1,0},{0,-1},{-1,0}};int mov 0; //用于确定移动方向while(mov 4){//此次探索位置的行坐标和列坐标int nextX curX direction[mov][0];int nextY curY direction[mov][1];//探索到的位置就是出口if(nextX exitX nextY exitY){//将当前坐标入栈为了打印路径push(LinkStack_X,curX);push(LinkStack_Y,curY);//打印出口坐标printf(深度优先搜索的路径\n(%d,%d),exitX,exitY);//反向打印路径坐标依次出栈while(!isEmpty(LinkStack_X)){curX pop(LinkStack_X);curY pop(LinkStack_Y);printf( - (%d,%d),curX,curY);}return 1;}//此次探访到的坐标不是出口但它可以到达且尚未访问else if(maze[nextX][nextY] 0 mark[nextX][nextY] 0){//此次探索的坐标标记为已访问mark[nextX][nextY]1;//当前坐标入栈作为路径的一部分栈中的路径结点不包含出口push(LinkStack_X,curX);push(LinkStack_Y,curY);//更新当前坐标curX nextX;curY nextY;//新的当前坐标重新从东开始探索即深度优先mov0;}else //遇墙或者该探访坐标已访问改变探索方向mov;}//4个方向都探索完了死胡同。//退回大循环通过出栈返回到上一个坐标以这个坐标为基坐标重新进行探索还不行就继续返回}//所有坐标的所有方向都探索完了即所有路径都探索完了可判断没有从入口到出口的通路return 0;
}int main()
{fileRead(); //从文件读取迷宫//将入口设置在左上角出口设置在右下脚。int entryX 1;int entryY 1;int exitX 7;int exitY 7;printfMaze(); //打印迷宫//搜索一条从入口到出口的路径int find mazeDFS(entryX,entryY,exitX,exitY);/* if(find 1)printfMaze(); //打印迷宫和路径elseprintf(找不到路径);*/return 0;
} 测试用例 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 输出结果 第2题
算法思想
使用深度优先搜索DFS从起始位置开始在每个位置上探索所有可能的路径。
使用递归通过递归调用在每个位置上继续搜索下一个位置。
使用回溯在搜索过程中如果无法继续前进则回溯到上一个位置并尝试其他路 算法实现
// Created by Mr.Chen on 2023/12/25.#includestdio.h
#includestdlib.h/*** 9*9迷宫(最外围是一堵墙)* 1代表墙0代表空格* 只有东西南北4个方向可以走*/
#define SIZE 9 //迷宫大小
int maze[SIZE][SIZE]; //迷宫/*** 存储路径坐标的结点称为路径结点* 因为数量无法确定故使用链式结构*/
typedef struct Node
{int coordinate; //坐标struct Node* next;
}Node;typedef Node* PNode; //路径结点的指针类型
typedef Node* LinkStack; //链栈指针,可代表栈顶指针/*** 创建一个链栈并返回栈顶指针*/
LinkStack creatLinkStack()
{LinkStack top (LinkStack)malloc(sizeof(Node));if(top ! NULL)top-next NULL;elseprintf(链栈创建失败);return top;
}/*** param top 坐标要入的栈的栈顶指针* param coordinate 入栈坐标*/
void push(LinkStack top, int coordinate)
{PNode pNode (PNode)malloc(sizeof(Node));if(pNode NULL)printf(入栈失败);else{pNode-coordinate coordinate;//头插法pNode-next top-next;top-next pNode;}
}/*** 空栈返回1非空返回0*/
int isEmpty(LinkStack top)
{return top-nextNULL ? 1 : 0;
}/*** 弹出栈顶元素的坐标*/
int pop(LinkStack top)
{int coordinate;coordinate top-next-coordinate; //获取栈顶元素的坐标//删除栈顶元素更新栈顶指针PNode t top-next;top-next t-next;free(t);return coordinate;
}/*** 从文件中读取迷宫*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\1.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);
}/*** 打印迷宫*/
void printfMaze()
{printf(迷宫■(墙) .(路) ♧(入口) ♤(出口)\n);printf( );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{/*if(path[i][j] 1)printf(√ );else*/printf(. );}}printf(\n);}}/*** 递归深度优先搜索打印所有路径* param entryX 入口行坐标* param entryY 入口列坐标* param exitX 出口行坐标* param exitY 出口列坐标* param maze 迷宫* param path 路径栈path也可以用作栈顶指针* param visited 访问标记数组*/
void mazeDFS(int entryX, int entryY, int exitX, int exitY, LinkStack path, int visited[SIZE][SIZE])
{//递归的返回条件到达终点。每返回一次打印一条路径if (entryX exitX entryY exitY){//打印出口printf((%d,%d), exitX, exitY);PNode p path-next;while (p ! NULL){//将栈中的元素解码得到路径上的坐标printf( - (%d,%d), p-coordinate / SIZE, p-coordinate % SIZE);p p-next;}printf(\n); //换行打印下一条路径printf(\n);return;}int direction[4][2]{{0,1},{1,0},{0,-1},{-1,0}};for (int mov 0; mov 4; mov){//此次探访的坐标int nextX entryX direction[mov][0];int nextY entryY direction[mov][1];//探访位置是空格且尚未访问if (maze[nextX][nextY] 0 visited[nextX][nextY] 0){//将本次起点的坐标编码后压入路径栈并将其标记为已访问push(path,entryX * SIZE entryY);visited[entryX][entryY] 1;//递归mazeDFS(nextX, nextY, exitX, exitY, path, visited);//每次搜索结束后都将上一个坐标标记为尚未未访问这是为了换下一个方向继续探索visited[entryX][entryY] 0;//在路径中删除该坐标pop(path);}}}int main()
{//从文件中读取迷宫fileRead();//将入口设置在左上角出口设置在右下脚int entryX 1;int entryY 1;int exitX 7;int exitY 7;//打印迷宫结构printfMaze();//创建路径栈LinkStack path creatLinkStack();//访问标记数组。不能在递归中初始化int visited[SIZE][SIZE] {0};//递归深搜打印所有路径printf(所有路径如下\n);mazeDFS(entryX,entryY,exitX,exitY,path,visited);return 0;
}
测试用例 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 输出结果 (7,7) - (7,6) - (6,6) - (6,5) - (6,4) - (6,3) - (5,3) - (4,3) - (3,3) - (2,3) - (2,2) - (1,2) - (1,1)
(7,7) - (7,6) - (6,6) - (6,5) - (6,4) - (6,3) - (5,3) - (4,3) - (3,3) - (2,3) - (2,2) - (2,1) - (1,1) 第3题 广度优先搜索
算法分析
广度优先搜索在迷宫问题中找出的第一条路径就是最短路径这是因为BFS从起点开始呈放射状扩散首先探索离起点最近的节点然后逐步向外扩展直到找到终点。因此只要从起点开始的扩散过程能够遍历到终点那么起点和终点之间一定是连通的他们之间至少存在一条路径。
具体来说广度优先搜索会按照队列中的元素顺序逐层访问迷宫中的节点。每一层的所有节点会被访问并标记为已访问然后才会访问下一层的节点。由于广度优先搜索是按照距离的递增顺序访问节点的所以它首先访问到的就是最短路径。
需要注意的是广度优先搜索假设迷宫中没有环路。如果存在环路那么广度优先搜索可能会陷入死循环无法找到正确的路径。
所以我们只要为迷宫的格子设置代价然后在第一题的广度优先搜索算法的基础上做一些更改即可
算法实现
// Created by Mr.Chen on 2023/12/25.
#include stdio.h
#include stdlib.h/*** 9*9迷宫(最外围是一堵墙)* 1代表墙0代表空格* 只有东西南北4个方向可以走*/
#define SIZE 9 //迷宫大小
int maze[SIZE][SIZE]; //迷宫
int path[SIZE][SIZE] {0}; //记录路径上的坐标标记值为1
int cost[SIZE][SIZE]; //每个格子的代价/*** 存储路径坐标的结点称为路径结点* 因为数量无法确定故使用链式结构*/
typedef struct Node
{int coordinate; //坐标int distance; //起点该结点的距离int totalCost; //起点到该结点的代价struct Node* next;
}Node;typedef Node* PNode; //路径结点的指针类型typedef struct Queue
{PNode front; //队头指针指向队头元素PNode rear; //队尾指针指向队尾元素
}Queue;typedef Queue* LinkQueue; //链队列/*** 创建并初始化一个链队列*/
LinkQueue creatLinkQueue()
{LinkQueue linkQueue (LinkQueue)malloc(sizeof(Queue));linkQueue-frontNULL;linkQueue-rearNULL;return linkQueue;
}/*** 空队返回1非空返回0*/
int isEmpty(LinkQueue linkQueue)
{return linkQueue-front NULL;
}/*** 入队*/
void inQueue(LinkQueue linkQueue,int coordinate, int distance, int cost)
{PNode pNode (PNode)malloc(sizeof(Node));//存储数据pNode-coordinate coordinate;pNode-distance distance;pNode-totalCost cost;pNode-next NULL;if(isEmpty(linkQueue)){linkQueue-front pNode;linkQueue-rear pNode;}else{//链队非空新的路径结点接在尾结点之后linkQueue-rear-next pNode;//更新尾指针linkQueue-rear pNode;}
}/*** 出队。返回并删除队头元素*/
int outQueue(LinkQueue linkQueue)
{int coordinate linkQueue-front-coordinate;PNode t linkQueue-front;linkQueue-frontlinkQueue-front-next; //改变指向跨越删除free(t);return coordinate;
}/*** 释放队列的内存*/
void freeQueue(LinkQueue linkQueue)
{PNode t (PNode)malloc(sizeof(Node));while(linkQueue-front-next ! NULL){t linkQueue-front-next;linkQueue-front-next linkQueue-front-next-next;free(t);linkQueue-front linkQueue-front-next;}free(linkQueue-front);
}/*** 从文件中读取迷宫和代价*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\1.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);FILE* fp2;/* 打开用于读取的文件路径不能有中文 */fp2 fopen(D:\\Projects\\ClionProjects\\mazeText\\3_cost.txt, r);if(fp2 NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp2,%d,cost[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp2,\n);}fclose(fp2);
}/*** 打印迷宫和路径*/
void printfMaze()
{printf(迷宫以及路径■(墙) .(路) ♧(入口) ♤(出口) √(路径)\n);printf( );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{if(path[i][j] 1)printf(√ );elseprintf(. );}}printf(\n);}//打印路径坐标path[1][1] 1;path[SIZE - 2][SIZE - 2] 1;for(int i 0; i SIZE -1; i){for(int j 0; j SIZE - 1; j)if(path[i][j] 1)printf(-(%d,%d), i, j);}
}void printCost()
{printf(将迷宫的代价设置为\n);for (int i 0; i SIZE - 1; i){for (int j 0; j SIZE - 1; j)printf(%d ,cost[i][j]);printf(\n);}printf(则);}/*** 广度优先搜索一条从入口到出口的路径,并反向打印输出* param entryX 入口行坐标* param entryY 入口列坐标* param exitX 出口行坐标* param exitY 出口列坐标* param maze 迷宫* return 存在路径返回1否则返回0*/
int mazeBSF(int entryX, int entryY, int exitX, int exitY)
{//迷宫的访问标记数组。未访问0 已访问1int mark[9][9] {0};//从入口开始探索故将入口标记为已访问mark[entryX][entryY] 1;//记录行走过程中前驱结点的坐标借助这些记录来回溯到起点int preX_record[SIZE][SIZE];int preY_record[SIZE][SIZE];//前驱结点的坐标初始化为-1for(int i 0; i SIZE; i){for(int j 0; j SIZE; j){preX_record[i][j]-1;preY_record[i][j]-1;}}//创建两个链队列//linkQueueX存储行坐标LinkQueue linkQueueX creatLinkQueue();//linkQueueX存储行坐标LinkQueue linkQueueY creatLinkQueue();//入口坐标入队//入口到入口的距离为0代价由代价数组确定出口和入口本身也有代价inQueue(linkQueueX, entryX,0,cost[entryX][entryY]);inQueue(linkQueueY, entryY,0,cost[entryX][entryY]);mark[entryX][entryY] 1;//还有路走就一直循环while(!isEmpty(linkQueueX)){/*** 这里可以优化* 1.让outQueue返回的是队头结点* 2.结点同时存储行坐标和列坐标这样只需要一个队列节省一倍空间*/int distance linkQueueX-front-distance;int totalCost linkQueueX-front-totalCost;//队头元素出队获取根据地坐标。可理解为正在探访的位置的前驱坐标int preX outQueue(linkQueueX);int preY outQueue(linkQueueY);//每次都绕根据地一圈进行探索for(int mov 0; mov 4 ; mov){//4个移动方向按东、南、西、北的顺时针顺序依次探索下一个位置int direction[4][2]{{0,1},{1,0},{0,-1},{-1,0}};//正在探索的位置的坐标int X preX direction[mov][0];int Y preY direction[mov][1];//找到出口if (X exitX Y exitY){//记录出口的前驱位置preX_record[X][Y] preX;preY_record[X][Y] preY;//回溯while(X ! entryX Y ! entryY){preX preX_record[X][Y];preY preY_record[X][Y];path[preX][preY] 1; //记录路径X preX;Y preY;}//打印该路径上入口到出口的距离和代价printf(入口到出口的最短距离%d\n,distance 1);printf(该路径的代价%d\n,totalCost cost[exitX][exitY]);return 1;}//此次探访到的坐标不是出口但它可以到达且尚未访问else if(maze[X][Y] 0 mark[X][Y] 0){//此次探访到的结点入队作为未来的根据地inQueue(linkQueueX,X,distance 1,totalCost cost[X][Y]);inQueue(linkQueueY,Y,distance 1,totalCost cost[X][Y]);//标记为已访问mark[X][Y]1;//记录此次探访结点的前驱preX_record[X][Y] preX;preY_record[X][Y] preY;}}}return 0;
}int main()
{fileRead(); //从文件读取迷宫和代价//打印代价printCost();//将入口设置在左上角出口设置在右下脚。int entryX 1;int entryY 1;int exitX 7;int exitY 7;//广度优先搜索一条最短路径int find mazeBSF(entryX,entryY,exitX,exitY);if(find 1)printfMaze(); //打印迷宫和路径elseprintf(找不到路径);return 0;
}
测试用例 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 迷宫代价 7 4 2 9 1 3 5 8 6 0 3 8 5 6 2 4 7 9 9 6 1 3 7 4 8 2 5 2 8 4 6 0 9 1 7 3 5 1 7 3 8 4 6 9 2 4 9 6 2 5 7 8 1 3 3 2 5 8 9 1 7 4 6 6 7 9 1 2 5 3 8 0 0 5 3 7 8 9 2 4 6 输出结果 A*算法
逻辑和第一题所用的算法基本相同
算法实现
// Created by Mr.Chen on 2023/12/29.
#include stdio.h
#include stdlib.h
#include stdbool.h#define SIZE 9
int maze[SIZE][SIZE];
int cost[SIZE][SIZE]; //每个格子的代价
bool visited[SIZE][SIZE] {false}; //访问标记数组
int path[SIZE][SIZE] {0};// 定义迷宫的起点和终点
typedef struct {int x;int y;
} coordinate;coordinate start {1, 1};
coordinate end {SIZE-2, SIZE-2};// 定义优先级队列的节点
typedef struct {int f; // f(n) g(n) h(n)int g; // 从起点到当前节点的路径长度coordinate point; // 当前节点的坐标
} Node;// 比较两个节点的优先级
int compare(const void *a, const void *b) {return ((Node*)a)-f - ((Node*)b)-f;
}// 获取一个current节点的所有邻居节点
void getNeighbors(coordinate current, coordinate neighbors[]) {int idx 0;if (current.x 1) {neighbors[idx].x current.x - 1;neighbors[idx].y current.y;idx;}if (current.x SIZE - 2) {neighbors[idx].x current.x 1;neighbors[idx].y current.y;idx;}if (current.y 1) {neighbors[idx].x current.x;neighbors[idx].y current.y - 1;idx;}if (current.y SIZE - 2) {neighbors[idx].x current.x;neighbors[idx].y current.y 1;}
}// 计算启发函数曼哈顿距离
int heuristic(coordinate node, coordinate goal) {return abs(node.x - goal.x) abs(node.y - goal.y);
}// 使用A*算法求解最短路径
int Astar() {//初始化存储结点的队列Node queue[SIZE * SIZE];int front 0, rear 0;//初始化各结点到起点的距离表默认为无穷大int distances[SIZE][SIZE];for (int i 0; i SIZE; i) {for (int j 0; j SIZE; j) {distances[i][j] INT_MAX;}}distances[start.x][start.y] cost[start.x][start.y];//记录行走过程中前驱结点的坐标借助这些记录来回溯到起点int preX_record[SIZE][SIZE];int preY_record[SIZE][SIZE];//前驱结点的坐标初始化为-1for(int i 0; i SIZE; i){for(int j 0; j SIZE; j){preX_record[i][j]-1;preY_record[i][j]-1;}}// 创建起点节点计算启发函数并加入队列Node startNode {heuristic(start, end), cost[start.x][start.y],start};queue[rear] startNode;// 开始执行A*算法while (front ! rear) {// 从队列中取出优先级最高的节点qsort(queue front, rear - front, sizeof(Node), compare);Node current queue[front];// 如果当前节点是终点返回最短路径长度if (current.point.x end.x current.point.y end.y){//回溯while(current.point.x ! start.x current.point.y ! start.y){int preX preX_record[current.point.x][current.point.y];int preY preY_record[current.point.x][current.point.y];path[preX][preY] 1; //记录路径current.point.x preX;current.point.y preY;}return current.g;}//获取当前节点的所有邻居节点coordinate neighbors[4];getNeighbors(current.point, neighbors);for (int i 0; i 4; i) {coordinate neighbor neighbors[i];if (maze[current.point.x][current.point.y] 0 visited[neighbor.x][neighbor.y] false) {visited[neighbor.x][neighbor.y] true;// 计算新的路径长度int newDist current.g cost[neighbor.x][neighbor.y];// 如果新的路径更短则更新距离表并加入队列if (newDist distances[neighbor.x][neighbor.y]) {distances[neighbor.x][neighbor.y] newDist;Node newNode {newDist heuristic(neighbor, end), newDist, neighbor};queue[rear] newNode;//记录此次探访结点的前驱preX_record[neighbor.x][neighbor.y] current.point.x;preY_record[neighbor.x][neighbor.y] current.point.y;}}}}// 没有找到最短路径返回-1表示失败return -1;
}/*** 从文件中读取迷宫和代价*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\1.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);FILE* fp2;/* 打开用于读取的文件路径不能有中文 */fp2 fopen(D:\\Projects\\ClionProjects\\mazeText\\3_cost.txt, r);if(fp2 NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp2,%d,cost[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp2,\n);}fclose(fp2);
}/*** 打印迷宫和路径*/
void printfMaze()
{printf(迷宫■(墙) .(路) ♧(入口) ♤(出口) √(路径)\n);printf( );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{if(path[i][j] 1)printf(√ );elseprintf(. );}}printf(\n);}//打印路径坐标printf(从入口到出口的代价最小的路径\n);path[1][1] 1;path[SIZE - 2][SIZE - 2] 1;for(int i 0; i SIZE -1; i){for(int j 0; j SIZE - 1; j)if(path[i][j] 1)printf(-(%d,%d), i, j);}printf(\n);
}void printCost()
{printf(将迷宫的代价设置为\n);for (int i 0; i SIZE - 1; i){for (int j 0; j SIZE - 1; j)printf(%d ,cost[i][j]);printf(\n);}printf(\n);
}int main()
{fileRead(); //从文件读取迷宫和代价//打印代价printCost();int shortestPathLength Astar();printfMaze();printf(代价为: %d\n, shortestPathLength);return 0;
} 测试用例 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 迷宫代价 7 4 2 9 1 3 5 8 6 0 3 8 5 6 2 4 7 9 9 6 1 3 7 4 8 2 5 2 8 4 6 0 9 1 7 3 5 1 7 3 8 4 6 9 2 4 9 6 2 5 7 8 1 3 3 2 5 8 9 1 7 4 6 6 7 9 1 2 5 3 8 0 0 5 3 7 8 9 2 4 6 输出结果 第4题
笔者暂时无法随机生成满足要求的迷宫只能自行设置一个90*90的迷宫。读者可自行实现随机生成迷宫
A*算法
算法实现
// Created by Mr.Chen on 2023/12/29.
#include stdio.h
#include stdlib.h
#include time.h
#include stdbool.h#define SIZE 90
int maze[SIZE][SIZE];
int cost[SIZE][SIZE]; //每个格子的代价
bool visited[SIZE][SIZE] {false}; //访问标记数组
int path[SIZE][SIZE] {0};// 定义迷宫的起点和终点
typedef struct {int x;int y;
} coordinate;coordinate start {1, 1};
coordinate end {SIZE-2, SIZE-2};// 定义优先级队列的节点
typedef struct {int f; // f(n) g(n) h(n)int g; // 从起点到当前节点的路径长度coordinate point; // 当前节点的坐标
} Node;// 比较两个节点的优先级
int compare(const void *a, const void *b) {return ((Node*)a)-f - ((Node*)b)-f;
}// 获取一个current节点的所有邻居节点
void getNeighbors(coordinate current, coordinate neighbors[]) {int idx 0;if (current.x 1) {neighbors[idx].x current.x - 1;neighbors[idx].y current.y;idx;}if (current.x SIZE - 2) {neighbors[idx].x current.x 1;neighbors[idx].y current.y;idx;}if (current.y 1) {neighbors[idx].x current.x;neighbors[idx].y current.y - 1;idx;}if (current.y SIZE - 2) {neighbors[idx].x current.x;neighbors[idx].y current.y 1;}
}// 计算启发函数曼哈顿距离
int heuristic(coordinate node, coordinate goal) {return abs(node.x - goal.x) abs(node.y - goal.y);
}// 使用A*算法求解最短路径
int Astar() {//初始化存储结点的队列Node queue[SIZE * SIZE];int front 0, rear 0;//初始化各结点到起点的距离表默认为无穷大int distances[SIZE][SIZE];for (int i 0; i SIZE; i) {for (int j 0; j SIZE; j) {distances[i][j] INT_MAX;}}distances[start.x][start.y] cost[start.x][start.y];//记录行走过程中前驱结点的坐标借助这些记录来回溯到起点int preX_record[SIZE][SIZE];int preY_record[SIZE][SIZE];//前驱结点的坐标初始化为-1for(int i 0; i SIZE; i){for(int j 0; j SIZE; j){preX_record[i][j]-1;preY_record[i][j]-1;}}// 创建起点节点计算启发函数并加入队列Node startNode {heuristic(start, end), cost[start.x][start.y],start};queue[rear] startNode;// 开始执行A*算法while (front ! rear) {// 从队列中取出优先级最高的节点qsort(queue front, rear - front, sizeof(Node), compare);Node current queue[front];// 如果当前节点是终点返回最短路径长度if (current.point.x end.x current.point.y end.y){//回溯while(current.point.x ! start.x current.point.y ! start.y){int preX preX_record[current.point.x][current.point.y];int preY preY_record[current.point.x][current.point.y];path[preX][preY] 1; //记录路径current.point.x preX;current.point.y preY;}return current.g;}//获取当前节点的所有邻居节点coordinate neighbors[4];getNeighbors(current.point, neighbors);for (int i 0; i 4; i) {coordinate neighbor neighbors[i];if (maze[current.point.x][current.point.y] 0 visited[neighbor.x][neighbor.y] false) {visited[neighbor.x][neighbor.y] true;// 计算新的路径长度int newDist current.g cost[neighbor.x][neighbor.y];// 如果新的路径更短则更新距离表并加入队列if (newDist distances[neighbor.x][neighbor.y]) {distances[neighbor.x][neighbor.y] newDist;Node newNode {newDist heuristic(neighbor, end), newDist, neighbor};queue[rear] newNode;//记录此次探访结点的前驱preX_record[neighbor.x][neighbor.y] current.point.x;preY_record[neighbor.x][neighbor.y] current.point.y;}}}}// 没有找到最短路径返回-1表示失败return -1;
}/*** 从文件中读取迷宫和代价*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\90.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);FILE* fp2;/* 打开用于读取的文件路径不能有中文 */fp2 fopen(D:\\Projects\\ClionProjects\\mazeText\\90_cost.txt, r);if(fp2 NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp2,%d,cost[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp2,\n);}fclose(fp2);
}/*** 打印迷宫和路径*/
void printfMaze()
{printf(迷宫■(墙) .(路) ♧(入口) ♤(出口) √(路径)\n);printf(\n );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路else{if(path[i][j] 1)printf(√ );elseprintf(. );}}printf(\n);}//打印路径坐标printf(代价最小的路径\n);path[1][1] 1;path[SIZE - 2][SIZE - 2] 1;for(int i 0; i SIZE -1; i){for(int j 0; j SIZE - 1; j)if(path[i][j] 1)printf(-(%d,%d), i, j);}printf(\n);
}void printCost()
{printf(将迷宫的代价设置为\n);for (int i 0; i SIZE - 1; i){for (int j 0; j SIZE - 1; j)printf(%d ,cost[i][j]);printf(\n);}printf(\n);
}int main()
{fileRead(); //从文件读取迷宫和代价//打印代价printCost();int begintime,endtime;begintimeclock(); //计时开始int shortestPathLength Astar();endtime clock(); //计时结束printfMaze();printf(代价为: %d\n, shortestPathLength);printf(耗时%d毫秒,endtime-begintime);return 0;
}
测试用例
由于字数过多这里仅给出迷宫代价请读者自行生成 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 0 0 1 0 0 0 1 1 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 输出结果 -(1,1)-(2,1)-(2,2)-(3,2)-(3,3)-(3,4)-(4,4)-(5,4)-(5,5)-(6,5)-(6,6)-(7,6)-(8,6)-(9,6)-(10,6)-(11,6)-(12,6)-(12,7)-(12,8)-(13,8)-(14,6)-(14,7)-(14,8)-(15,6)-(16,6)-(17,6)-(18,6)-(19,6)-(20,6)-(20,7)-(21,7)-(22,7)-(23,7)-(23,8)-(24,8)-(25,8)-(25,9)-(26,9)-(27,9)-(28,6)-(28,7)-(28,8)-(28,9)-(29,6)-(30,6)-(31,6)-(31,7)-(31,8)-(31,9)-(32,9)-(32,10)-(33,10)-(34,10)-(34,11)-(35,11)-(36,11)-(37,11)-(38,11)-(39,11)-(40,10)-(40,11)-(41,10)-(42,10)-(43,10)-(44,10)-(45,9)-(45,10)-(46,9)-(47,9)-(47,10)-(47,11)-(47,12)-(47,13)-(48,13)-(48,14)-(49,14)-(49,15)-(50,15)-(50,16)-(51,16)-(52,16)-(53,16)-(54,16)-(55,16)-(56,11)-(56,12)-(56,13)-(56,14)-(56,15)-(56,16)-(57,11)-(58,11)-(58,12)-(59,12)-(59,13)-(60,13)-(61,13)-(61,14)-(62,14)-(62,15)-(62,16)-(63,16)-(63,17)-(64,17)-(64,18)-(65,18)-(66,18)-(67,18)-(67,19)-(67,20)-(67,21)-(67,22)-(68,22)-(68,23)-(69,23)-(69,24)-(69,25)-(70,25)-(70,26)-(71,26)-(71,27)-(72,27)-(72,28)-(73,28)-(73,29)-(74,29)-(75,29)-(75,30)-(76,30)-(76,31)-(76,32)-(76,33)-(76,34)-(76,35)-(76,36)-(76,37)-(77,37)-(78,37)-(78,38)-(78,41)-(78,42)-(78,43)-(78,44)-(78,45)-(78,46)-(78,47)-(78,48)-(78,49)-(78,50)-(79,38)-(79,39)-(79,40)-(79,41)-(79,50)-(80,50)-(80,51)-(80,52)-(80,53)-(80,54)-(80,55)-(80,56)-(81,56)-(82,56)-(82,57)-(82,58)-(82,59)-(83,59)-(83,60)-(83,61)-(84,61)-(84,62)-(84,63)-(84,64)-(84,65)-(85,65)-(85,66)-(85,67)-(85,68)-(85,69)-(86,69)-(86,70)-(86,71)-(86,72)-(86,81)-(86,82)-(86,83)-(86,84)-(86,85)-(86,86)-(86,87)-(86,88)-(87,72)-(87,73)-(87,74)-(87,76)-(87,77)-(87,78)-(87,79)-(87,81)-(87,88)-(88,74)-(88,75)-(88,76)-(88,79)-(88,80)-(88,81)-(88,88) 深度优先搜索
介绍
这个算法比较暴力且粗糙且没有很好地完成题目要求。我们只能在第2题“打印所有路径”的基础上动态维护最小的路径代价并没有能直接记录最小代价路径还得在输出结果中人工搜索。所以这是未来笔者的一个优化点甚至考虑使用另外一种算法。
算法实现
// Created by Mr.Chen on 2023/12/27.#includestdio.h
#includestdlib.h
#include time.h/*** 迷宫(最外围是一堵墙)* 1代表墙0代表空格* 只有东西南北4个方向可以走*/
#define SIZE 90 //迷宫大小
int maze[SIZE][SIZE];
int cost[SIZE][SIZE]; //每个格子的代价
int minCost INT_MAX; //最小路径代价
/*** 存储路径坐标的结点称为路径结点* 因为数量无法确定故使用链式结构*/
typedef struct Node
{int coordinate; //坐标struct Node* next;
}Node;typedef Node* PNode; //路径结点的指针类型
typedef Node* LinkStack; //链栈指针,可代表栈顶指针/*** 创建一个链栈并返回栈顶指针*/
LinkStack creatLinkStack()
{LinkStack top (LinkStack)malloc(sizeof(Node));if(top ! NULL)top-next NULL;elseprintf(链栈创建失败);return top;
}/*** param top 坐标要入的栈的栈顶指针* param coordinate 入栈坐标*/
void push(LinkStack top, int coordinate)
{PNode pNode (PNode)malloc(sizeof(Node));if(pNode NULL)printf(入栈失败);else{pNode-coordinate coordinate;//头插法pNode-next top-next;top-next pNode;}
}/*** 空栈返回1非空返回0*/
int isEmpty(LinkStack top)
{return top-nextNULL ? 1 : 0;
}/*** 弹出栈顶元素的坐标*/
int pop(LinkStack top)
{int coordinate;coordinate top-next-coordinate; //获取栈顶元素的坐标//删除栈顶元素更新栈顶指针PNode t top-next;top-next t-next;free(t);return coordinate;
}/*** 从文件中读取迷宫和代价*/
void fileRead()
{FILE* fp;/* 打开用于读取的文件路径不能有中文 */fp fopen(D:\\Projects\\ClionProjects\\mazeText\\90.txt, r);if(fp NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp,%d,maze[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp,\n);}fclose(fp);FILE* fp2;/* 打开用于读取的文件路径不能有中文 */fp2 fopen(D:\\Projects\\ClionProjects\\mazeText\\90_cost.txt, r);if(fp2 NULL){perror(打开文件时发生错误);return;}for(int i0; i SIZE; i){for(int j 0;j SIZE;j)fscanf(fp2,%d,cost[i][j]);/*每次读取一个数fscanf函数遇到空格或者换行结束*/fscanf(fp2,\n);}fclose(fp2);
}/*** 打印迷宫*/
void printfMaze()
{printf(迷宫■(墙) .(路) ♧(入口) ♤(出口)\n);printf(\n );for(int i 0; i SIZE; i)printf(%d ,i);printf(\n);for(int i 0; i SIZE; i){//可视化路径printf(%d ,i);for(int j 0; j SIZE; j){//打印入口if(i 1 j 1)printf(♧ );//打印出口else if(i SIZE - 2 j SIZE - 2)printf(♤ );//打印墙else if(maze[i][j] 1)printf(■ );//打印路elseprintf(. );}printf(\n);}
}void printCost()
{printf(将迷宫的代价设置为\n);for (int i 0; i SIZE - 1; i){for (int j 0; j SIZE - 1; j)printf(%d ,cost[i][j]);printf(\n);}printf(则);}/*** 递归深度优先搜索打印所有路径* param entryX 入口行坐标* param entryY 入口列坐标* param exitX 出口行坐标* param exitY 出口列坐标* param maze 迷宫* param path 路径栈path也可以用作栈顶指针* param visited 访问标记数组*/
void mazeDFS(int entryX, int entryY, int exitX, int exitY, LinkStack path, int visited[SIZE][SIZE], int totalCost)
{//递归的返回条件到达终点。每返回一次打印一条路径if (entryX exitX entryY exitY){if(totalCost minCost)minCost totalCost; //维护最小的路径代价//打印出口printf(该路径的代价%d。(%d,%d), totalCost, exitX, exitY);PNode p path-next;while (p ! NULL){//将栈中的元素解码得到路径上的坐标printf( - (%d,%d), p-coordinate / SIZE, p-coordinate % SIZE);p p-next;}printf(\n); //换行打印下一条路径printf(\n);return;}int direction[4][2]{{0,1},{1,0},{0,-1},{-1,0}};for (int mov 0; mov 4; mov){//此次探访的坐标int nextX entryX direction[mov][0];int nextY entryY direction[mov][1];//探访位置是空格且尚未访问if (maze[nextX][nextY] 0 visited[nextX][nextY] 0){//将本次起点的坐标编码后压入路径栈并将其标记为已访问push(path,entryX * SIZE entryY);visited[entryX][entryY] 1;//递归mazeDFS(nextX, nextY, exitX, exitY, path, visited, totalCost cost[nextX][nextY]);//每次搜索结束后都将上一个坐标标记为尚未未访问这是为了换下一个方向继续探索visited[entryX][entryY] 0;//在路径中删除该坐标pop(path);}}}int main()
{fileRead(); //从文件读取迷宫和代价//打印代价printCost();//将入口设置在左上角出口设置在右下脚int entryX 1;int entryY 1;int exitX SIZE - 2;int exitY SIZE - 2;//打印迷宫结构printfMaze();//创建路径栈LinkStack path creatLinkStack();//访问标记数组。不能在递归中初始化int visited[SIZE][SIZE] {0};//递归深搜printf(所有路径如下\n);int begintime,endtime;begintimeclock(); //计时开始mazeDFS(entryX,entryY,exitX,exitY,path,visited,cost[entryX][entryY]);endtime clock(); //计时结束printf(最小的路径代价%d\n,minCost);printf(耗时%d毫秒,endtime-begintime);return 0;
}
测试用例
同上
输出结果 效率分析
1. 时间效率 - DFS在最坏的情况下DFS需要遍历迷宫中的所有可能路径因此其时间复杂度为O(VE)其中V是顶点的数量E是边的数量。对于90*90的迷宫如果每个格子都是一个顶点那么总共有8100个顶点和72900条边所以DFS的时间复杂度为O(810072900)O(74000)。在这个迷宫大小为90x90的例子中A*算法的时间效率0毫秒优于深度优先搜索19毫秒一个数量级。 - A*算法A*算法使用启发式函数来估计从当前节点到目标节点的最短距离因此在找到最短路径时通常比DFS更快。对于90*90的迷宫A*算法的时间复杂度也是O(VE)但由于启发式函数的存在它通常会比DFS更快地找到最短路径。
2. 空间效率 - DFSDFS使用递归或显式栈来实现因此其空间复杂度为O(V)其中V是顶点的数量。对于90*90的迷宫DFS的空间复杂度为O(8100)。 - A*算法A*算法使用优先级队列来存储待处理的节点因此其空间复杂度也为O(V)。对于90*90的迷宫A*算法的空间复杂度为O(8100)。
总的来说A*算法在时间效率上通常优于DFS因为它可以更快地找到最短路径。然而在空间效率上两者相同因为它们都需要存储所有的顶点信息。 第5题
笔者暂时无法随机生成满足要求的迷宫只能自行设置一个180*180的迷宫。读者可自行实现随机生成迷宫。
仍然是第4题的两种算法我们将迷宫的大小修改为180*180发现两者时间效率的差距变得更明显。
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1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0 0 0 0 1 0 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 输出结果
A*算法 深度优先搜索 总结
实验中采用了三种常见的迷宫问题解决算法深度优先搜索DFS、广度优先搜索BFS和A*算法旨在对不同的迷宫问题解决算法进行评估和比较
我们设计了一个标准的迷宫问题并生成了不同大小的迷宫。然后我们分别使用DFS、BFS和A*算法对这些迷宫进行搜索并记录了每种算法的搜索时间和找到出口的概率。
实验结果显示DFS算法在小规模迷宫问题上表现出较好的性能搜索时间较短找到出口的概率较高。然而随着迷宫规模的增大DFS算法的性能下降明显搜索时间急剧增加找到出口的概率也大幅下降。
相比之下BFS算法在大规模迷宫问题上表现出较好的性能。无论是搜索时间还是找到出口的概率BFS算法都相对稳定且较为优秀。这表明BFS算法在处理复杂迷宫问题时具有更好的可扩展性和效率。缺点是开销较大无用搜索过多——这点可以通过A*算法改进。
A*算法在搜索时间和找到出口的概率上均表现出最佳性能。无论是小规模还是大规模迷宫问题A*算法都能在较短的时间内找到最短路径并成功出口。这表明A*算法在解决迷宫问题时具有更高的智能性和优化能力不愧是人工智能领域的一种重要算法。
综上所述本次对比实验研究表明在不同的迷宫问题规模下DFS、BFS和A*算法各有优劣。对于小规模迷宫问题DFS算法是一个较好的选择而对于大规模迷宫问题BFS和A*算法则更为适合。因此在选择迷宫问题解决算法时需要根据具体情况综合考虑算法的性能和适用性。 特别鸣谢
Back迷宫问题的对比实验研究
以及我的辛苦付出的队友
