网站架构设计师有哪些学校可以报考,深圳手机集团网站建设,app设计策划书,wordpress mingleB. Connecting Universities 大意: 给定树, 给定2*k个点, 求将2*k个点两两匹配, 每个匹配的贡献为两点的距离, 求贡献最大值 单独考虑每条边$(u,v)$的贡献即可, 最大贡献显然是左右两侧点的最小值. #include iostream
#include algorithm
#include cstdioiostream
#include algorithm
#include cstdio
#include math.h
#include set
#include map
#include queue
#include string
#include string.h
#include bitset
#define REP(i,a,n) for(int ia;in;i)
#define PER(i,a,n) for(int in;ia;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o1)
#define rc (lc|1)
#define mid ((lr)1)
#define ls lc,l,mid
#define rs rc,mid1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl \n
using namespace std;
typedef long long ll;
typedef pairint,int pii;
const int P 1e97, INF 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r1%P;for (a%P;n;aa*a%P,n1)if(n1)rr*a%P;return r;}
ll inv(ll x){return x1?1:inv(P%x)*(P-P/x)%P;}
//headconst int N 1e610;
int n, k;
int a[N], sz[N];
vectorint g[N];
ll ans;void dfs(int x, int fa) {sz[x] a[x];for (int y:g[x]) if (y!fa) { dfs(y,x), sz[x]sz[y];ans min(sz[y], k-sz[y]);}
}int main() {scanf(%d%d, n, k),k*2;REP(i,1,k) { int t;scanf(%d, t);a[t] 1;}REP(i,2,n) {int u, v;scanf(%d%d, u, v);g[u].pb(v),g[v].pb(u);}dfs(1,0);printf(%lld\n, ans);
} C. Break Up 大意: 无向有权图有重边自环, 求删除两条边使得s与t不连通, 且两条边的边权和最小. 先求出任意一条最短路径, 边数显然不超过$n$, 暴力枚举这$n$条边然后再tarjan即可, 复杂度O(n(mn)) 算是挺简单的了, 还是打了好久, 一直卡在怎么判断删除一条边后是否连通, 后来发现tarjan后从s-t经过的桥一定是一条链, 所以直接dfs就好了, 最后还要注意边权1e91e9爆掉0x3f3f3f3f了. #include iostream
#include algorithm
#include cstdio
#include math.h
#include set
#include map
#include queue
#include string
#include string.h
#include bitset
#define REP(i,a,n) for(int ia;in;i)
#define PER(i,a,n) for(int in;ia;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o1)
#define rc (lc|1)
#define mid ((lr)1)
#define ls lc,l,mid
#define rs rc,mid1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl \n
using namespace std;
typedef long long ll;
typedef pairint,int pii;
const int P 1e97, INF ~0u1;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r1%P;for (a%P;n;aa*a%P,n1)if(n1)rr*a%P;return r;}
ll inv(ll x){return x1?1:inv(P%x)*(P-P/x)%P;}
//headconst int N 3e410;
int n, m, S, T;
int w[N];
struct _ {int to,id;} fa[N];
vector_ g[N];
int dfn[N], low[N], isbridge[N], clk;
void tarjan(int x, int fa, int z) {dfn[x]low[x]clk;for (auto e:g[x]) if (e.id!z) {int y e.to, id e.id;if (!dfn[y]) {tarjan(y,id,z);low[x]min(low[x],low[y]);if (low[e.to]dfn[x]) isbridge[id]1;} else if (dfn[y]dfn[x]id!fa) { low[x]min(low[x],dfn[y]);}}
}
int vis[N], c[N];
int dfs(int x) {if (xT) return 1;for (auto e:g[x]) if (!vis[e.id]) {vis[e.id] 1;if (dfs(e.to)) return c[e.id] 1;}return 0;
}int main() {scanf(%d%d%d%d, n, m, S, T);REP(i,1,m) { int u, v;scanf(%d%d%d, u, v, wi);g[u].pb({v,i}), g[v].pb({u,i});}queueint q;fa[S].to-1, q.push(S);while (q.size()) {int x q.front(); q.pop();for (auto e:g[x]) if (!fa[e.to].to) {fa[e.to]{x,e.id}, q.push(e.to);}}if (!fa[T].to) return puts(0\n0),0;int ans INF;vectorint vec;for (int xT; x!S; xfa[x].to) {int id fa[x].id;memset(vis,0,sizeof vis);memset(c,0,sizeof c);vis[id] 1;if (!dfs(S)) {if (answ[id]) ans w[id],vec.clear(),vec.pb(id);continue;}memset(dfn,0,sizeof dfn);memset(isbridge,0,sizeof isbridge);clk 0;tarjan(S,0,id);REP(i,1,m) if (c[i]isbridge[i]answ[id]w[i]) {answ[id]w[i];vec.clear();vec.pb(id), vec.pb(i);}}if (ansINF) return puts(-1),0;printf(%d\n%d\n, ans, int(vec.size()));for (int t:vec) printf(%d , t); hr;
} D. Huffman Coding on Segment 莫队一下, 然后将出现次数小于等于$\sqrt{n}$的暴力合, 其余的用堆合, 复杂度$O(m\sqrt{n}logn)$, 看了下最优解, 好像可以排序一下省去堆从而优化掉一个log #include iostream
#include algorithm
#include cstdio
#include math.h
#include set
#include map
#include queue
#include string
#include string.h
#include bitset
#define REP(i,a,n) for(int ia;in;i)
#define PER(i,a,n) for(int in;ia;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o1)
#define rc (lc|1)
#define mid ((lr)1)
#define ls lc,l,mid
#define rs rc,mid1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl \n
using namespace std;
typedef long long ll;
typedef pairint,int pii;
const int P 1e97, INF 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r1%P;for (a%P;n;aa*a%P,n1)if(n1)rr*a%P;return r;}
ll inv(ll x){return x1?1:inv(P%x)*(P-P/x)%P;}
//head#ifdef ONLINE_JUDGE
const int N 1e610;
#else
const int N 111;
#endifint n, m, sqn;
int blo[N], cnt[N], sum[N], s[N], a[N];
struct _ {int l,r,id;bool operator (const _ rhs) const {return blo[l]^blo[rhs.l]?lrhs.l:blo[l]1?rrhs.r:rrhs.r;}
} e[N];
ll ans[N];
vectorint q;void upd(int x, int d) {--sum[cnt[x]];cnt[x]d;sum[cnt[x]];
}ll calc() {ll ans 0;REP(i,1,sqn) s[i] sum[i];priority_queueint,vectorint,greaterint Q;int pre 0;REP(i,1,sqn) if (s[i]) {if (pre) {int x prei;ans x;if (xsqn) Q.push(x);else s[x];--s[i], pre 0;}if (s[i]1) --s[i], pre i;ans s[i]*i;if (i*2sqn) s[i*2]s[i]/2;else {REP(j,1,s[i]/2) Q.push(i*2);}}if (pre) Q.push(pre);for (auto i:q) if (cnt[i]sqn) Q.push(cnt[i]);while (Q.size()1) {int x Q.top(); Q.pop();x Q.top(); Q.pop();ans x, Q.push(x);}return ans;
}int main() {scanf(%d, n), sqn sqrt(n);REP(i,1,n) scanf(%d,ai),cnt[a[i]],blo[i]i/sqn;REP(i,1,N-1) if (cnt[i]sqn) q.pb(i);memset(cnt,0,sizeof cnt);scanf(%d, m);REP(i,1,m) scanf(%d%d,e[i].l,e[i].r),e[i].idi;sort(e1,e1m);int ql1,qr0;REP(i,1,m) {while (qle[i].l) upd(a[ql],-1);while (qre[i].r) upd(a[qr--],-1);while (qle[i].l) upd(a[--ql],1);while (qre[i].r) upd(a[qr],1);ans[e[i].id]calc();}REP(i,1,m) printf(%lld\n, ans[i]);
}E. Cool Slogans 后缀自动机还没学, 以后补了 转载于:https://www.cnblogs.com/uid001/p/10574977.html
