bzoj2229: [Zjoi2011]最小割（最小割树）

传送门

这题是用最小割树做的（不明白最小割树是什么的可以去看看这一题->这里）

有了最小割树就很简单了……点数那么少……每次跑出一个最大流就暴力搞一遍就好了

 //minamoto
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #define inf 0x3f3f3f3f
 using namespace std;
 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 char buf[<<],*p1=buf,*p2=buf;
 template<class T>inline bool cmin(T&a,const T&b){return a>b?a=b,:;}
 inline int read(){
     #define num ch-'0'
     char ch;bool flag=;int res;
     while(!isdigit(ch=getc()))
     (ch=='-')&&(flag=true);
     for(res=num;isdigit(ch=getc());res=res*+num);
     (flag)&&(res=-res);
     #undef num
     return res;
 }
 char sr[<<],z[];int C=-,Z;
 inline void Ot(){fwrite(sr,,C+,stdout),C=-;}
 inline void print(int x){
     if(C><<)Ot();if(x<)sr[++C]=,x=-x;
     while(z[++Z]=x%+,x/=);
     while(sr[++C]=z[Z],--Z);sr[++C]='\n';
 }
 const int N=,M=;
 int head[N],Next[M],ver[M],edge[M],ee[M],tot=;
 queue<int> q;
 inline void add(int u,int v,int e){
     ver[++tot]=v,Next[tot]=head[u],head[u]=tot,edge[tot]=e,ee[tot]=e;
     ver[++tot]=u,Next[tot]=head[v],head[v]=tot,edge[tot]=e,ee[tot]=e;
 }
 int cur[N],dep[N],n,m,S,T;
 bool bfs(){
     memset(dep,-,sizeof(dep));
     for(int i=;i<=n;++i) cur[i]=head[i];
     while(!q.empty()) q.pop();
     q.push(S),dep[S]=;
     while(!q.empty()){
         int u=q.front();q.pop();
         for(int i=head[u];i;i=Next[i])
         if(dep[ver[i]]<&&edge[i]){
             q.push(ver[i]),dep[ver[i]]=dep[u]+;
             if(ver[i]==T) return true;
         }
     }
     return false;
 }
 int dfs(int u,int limit){
     if(u==T||!limit) return limit;
     int flow=,f;
     for(int i=cur[u];i;cur[u]=i=Next[i]){
         int v=ver[i];
         if(dep[v]==dep[u]+&&(f=dfs(v,min(limit,edge[i])))){
             flow+=f,limit-=f;
             edge[i]-=f,edge[i^]+=f;
             if(!limit) break;
         }
     }
     if(!flow) dep[u]=-;
     return flow;
 }
 int dinic(){
     int flow=;
     while(bfs()) flow+=dfs(S,inf);
     return flow;
 }
 int id[N],tmp[N],ans[N][N];
 void solve(int L,int R){
     if(L==R) return;
     for(int i=;i<=tot;i+=) edge[i]=edge[i^]=ee[i];
     S=id[L],T=id[R];
     int flow=dinic();
     for(int i=;i<=n;++i)
     if(~dep[i])
     for(int j=;j<=n;++j)
     if(dep[j]<)
     cmin(ans[i][j],flow),cmin(ans[j][i],flow);
     int l=L,r=R;
     for(int i=L;i<=R;++i)
     if(~dep[id[i]]) tmp[l++]=id[i];
     else tmp[r--]=id[i];
     memcpy(id+L,tmp+L,sizeof(int)*(R-L+));
     solve(L,l-),solve(r+,R);
 }
 inline void init(){
     memset(head,,sizeof(head)),tot=,memset(ans,0x3f,sizeof(ans));
 }
 int main(){
     //freopen("testdata.in","r",stdin);
     int T=read();
     while(T--){
         init();
         n=read(),m=read();
         for(int i=;i<=m;++i){
             int u=read(),v=read(),e=read();add(u,v,e);
         }
         for(int i=;i<=n;++i) id[i]=i;
         solve(,n);
         int q=read();
         while(q--){
             int k=read(),res=;
             for(int i=;i<=n;++i)
             for(int j=i+;j<=n;++j)
             if(ans[i][j]<=k) ++res;
             print(res);
         }
         sr[++C]='\n';
     }
     Ot();
     return ;
 }