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我是题目啊
这个题一看到求最小费用就想最小费用最大流，然后也没看懂题。。。
其实是求去掉几个顶点使s到t不连通，很明显最小割，然后就是建图，由于是去掉顶点，所以将每个点拆成两个点，容量即为这个点的花费，然后是每两个有关系的点相连，为了保证最后是一个环，所以，最后的图：
代码：
#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <vector>
#include <queue>
using namespace std;
const int N = 20100;
const int INF = 0x3f3f3f3f;
struct edge
{
int to, cap, next;
} g[N*100];
int level[N], iter[N], head[N], que[N];
int n, m, cnt;
void add_edge(int v, int u, int cap)
{
g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
bool bfs(int s, int t)
{
memset(level, -1, sizeof level);
level[s] = 0;
int st = 0, en = 0;
que[0] = s;
while(st <= en)
{
int v = que[st++];
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[u] < 0)
{
level[u] = level[v] + 1;
que[++en] = u;
}
}
}
return level[t] == -1;
}
int dfs(int v, int t, int f)
{
if(v == t) return f;
int tm = f;
for(int &i = iter[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[v] < level[u])
{
int d = dfs(u, t, min(tm, g[i].cap));
g[i].cap -= d, g[i^1].cap += d, tm -= d;
if(tm == 0) break;
}
}
if(tm == f) level[v] = -1;
return f - tm;
}
int dinic(int s, int t)
{
int flow = 0, f;
while(true)
{
if(bfs(s, t)) return flow;
memcpy(iter, head, sizeof head);
flow += dfs(s, t, INF);
}
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
cnt = 0;
memset(head, -1, sizeof head);
int s,t;
scanf("%d%d",&s,&t);
int x;
for(int i = 1; i <= n; i++)
{
scanf("%d",&x);
add_edge(i,i+n,x);
}
t=t+n;
int a,b;
for(int j = 1; j <=m; j++)
{
scanf("%d%d",&a,&b);
add_edge(a+n, b,INF);
add_edge(b+n, a,INF);
}
printf("%d\n", dinic(s,t));
}
}