HDU 3491 Thieves

题意：给出一张无向图，每个点上都有一个权值，然后让你删掉权值之和尽量小的点，使得S到H不连通。

无向图带权点连通度问题。每个点拆点，容留为点权，对于原来图中的无向边(u,v)，拆成两条边(u',v,inf),(v',u,inf)，跑S'到T 的maxflow就可以了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 1<<30
#define maxn 210
#define maxm 30000
using namespace std;

int v[maxm],next[maxm],w[maxm];
int first[maxn],d[maxn],work[maxn],q[maxn];
int e,S,T,n,m;

void init(){
    e = 0;
    memset(first,-1,sizeof(first));
}

void add_edge(int a,int b,int c){
    v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++;
    v[e] = a;next[e] = first[b];w[e] = 0;first[b] = e++;
}

int bfs(){
    int rear = 0;
    memset(d,-1,sizeof(d));
    d[S] = 0;q[rear++] = S;
    for(int i = 0;i < rear;i++){
        for(int j = first[q[i]];j != -1;j = next[j])
            if(w[j] && d[v[j]] == -1){
                d[v[j]] = d[q[i]] + 1;
                q[rear++] = v[j];
                if(v[j] == T)   return 1;
            }
    }
    return 0;
}

int dfs(int cur,int a){
    if(cur == T)    return a;
    for(int &i = work[cur];i != -1;i = next[i]){
        if(w[i] && d[v[i]] == d[cur] + 1)
            if(int t = dfs(v[i],min(a,w[i]))){
                w[i] -= t;w[i^1] += t;
                return t;
            }
    }
    return 0;
}

int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(work,first,sizeof(first));
        while(int t = dfs(S,INF))   ans += t;
    }
    return ans;
}

int main()
{
    int kase;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d%d%d%d",&n,&m,&S,&T);
        S += n;
        init();
        for(int i = 1;i <= n;i++){
            int tmp;
            scanf("%d",&tmp);
            add_edge(i,i+n,tmp);
        }
        for(int i = 1;i <= m;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            add_edge(a+n,b,INF);
            add_edge(b+n,a,INF);
        }
        printf("%d\n",dinic());
    }
    return 0;
}