** Question chain ：**

http://www.lydsy.com/JudgeOnline/problem.php?id=2229

** Answer key ：**

First, take a look at this blog ：http://blog.csdn.net/jyxjyx27/article/details/42750833

Very good , We can have a simple perceptual knowledge of minimal cut trees .

It's very troublesome to find the minimum cut tree , And the number of points in this question is not large ,

So there is no need to construct a minimum cut tree , Just figure out all the ans[i][j]：i->j The minimum cut .

That is, divide and conquer , Find out n-1 A minimal cut ,

And after every minimal cut , Try to update the ans that will do .

Code ：

#include<queue>

#include<cstdio>

#include<cstring>

#include<iostream>

#define MAXN 200

#define MAXM 7000

#define INF 0x3f3f3f3f

#define filein(x) freopen(#x".in","r",stdin);

#define fileout(x) freopen(#x".out","w",stdout);

using namespace std;

struct Edge{

int to[MAXM],cap[MAXM],next[MAXM],head[MAXN],ent;

void Init(){

ent=2; memset(head,0,sizeof(head));

}

void Adde(int u,int v,int w){

to[ent]=v; cap[ent]=w; next[ent]=head[u]; head[u]=ent++;

to[ent]=u; cap[ent]=0; next[ent]=head[v]; head[v]=ent++;

}

void reset(){

for(int i=2;i<ent;i+=2){

cap[i]=cap[i^1]=(cap[i]+cap[i^1])/2;

}

}

}E;

bool mark[MAXN];

int a[MAXN],cur[MAXN],d[MAXN],ans[MAXN][MAXN];

int N,M,Q;

bool bfs(int S,int T){

static queue<int> q; int u,v;

memset(d,0,sizeof(d));

d[S]=1; q.push(S);

while(!q.empty()){

u=q.front(); q.pop();

for(int i=E.head[u];i;i=E.next[i]){

v=E.to[i];

if(d[v]||!E.cap[i]) continue;

d[v]=d[u]+1; q.push(v);

}

}

return d[T];

}

int dfs(int u,int reflow,const int &T){

if(u==T||!reflow) return reflow;

int flowout=0,f,v;

for(int &i=cur[u];i;i=E.next[i]){

v=E.to[i];

if(d[v]!=d[u]+1) continue;

f=dfs(v,min(reflow,E.cap[i]),T);

flowout+=f; E.cap[i^1]+=f;

reflow-=f; E.cap[i]-=f;

if(!reflow) break;

}

if(!flowout) d[u]=0;

return flowout;

}

int Dinic(int S,int T){

int flow=0;

while(bfs(S,T)){

memcpy(cur,E.head,sizeof(E.head));

flow+=dfs(S,INF,T);

}

return flow;

}

void dfs(int u){

mark[u]=1;

for(int i=E.head[u];i;i=E.next[i])

if(!mark[E.to[i]]&&E.cap[i]) dfs(E.to[i]);

}

void Partition(int l,int r){

static int tmp[MAXN],S,T,cut;

if(l>=r) return;

E.reset(); S=a[l]; T=a[r];

cut=Dinic(S,T);

memset(mark,0,sizeof(mark)); dfs(S);

for(int i=1;i<=N;i++)

for(int j=1;j<i;j++) if(mark[a[i]]^mark[a[j]])

ans[a[i]][a[j]]=ans[a[j]][a[i]]=min(ans[a[i]][a[j]],cut);

int L=l,R=r;

for(int i=l;i<=r;i++)

if(mark[a[i]]) tmp[L++]=a[i];

else tmp[R--]=a[i];

for(int i=l;i<L;i++) a[i]=tmp[i];

for(int i=r;i>R;i--) a[i]=tmp[i];

Partition(l,L-1);

Partition(R+1,r);

}

int main()

{

filein(mincut); fileout(mincut);

int T; scanf("%d",&T);

while(T--){

E.Init();

memset(ans,0x3f,sizeof(ans));

scanf("%d%d",&N,&M);

for(int i=1;i<=N;i++) a[i]=i;

for(int i=1,u,v,w;i<=M;i++){

scanf("%d%d%d",&u,&v,&w);

E.Adde(u,v,2*w);

}

E.reset(); Partition(1,N);

scanf("%d",&Q);

for(int k=1,c,num;k<=Q;k++){

scanf("%d",&c); num=0;

for(int i=1;i<=N;i++)

for(int j=1;j<i;j++)

if(ans[i][j]<=c) num++;

printf("%d\n",num);

}

puts("");

} return 0;

}

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