題目大意：一個城市中有些點，有一些雙向道路將這些點連線起來，每個點都有權值，求警察最少佔據的點的權值和使得從A點無法到達B點。

思路：最小點割集簽到題。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 80010
#define INF 0x3f3f3f3f
using namespace std;
#define min(a,b) ((a) < (b) ? (a):(b))

int points,edges;
int S,T;
int head[MAX],total = 1;
int next[MAX],aim[MAX],flow[MAX];

inline void Add(int x,int y,int f)
{
	next[++total] = head[x];
	aim[total] = y;
	flow[total] = f;
	head[x] = total;
}

int deep[MAX];

inline bool BFS()
{
	static queue<int> q;
	while(!q.empty())	q.pop();
	memset(deep,0,sizeof(deep));
	deep[S] = 1;
	q.push(S);
	while(!q.empty()) {
		int x = q.front(); q.pop();
		for(int i = head[x]; i; i = next[i])
			if(flow[i] && !deep[aim[i]]) {
				deep[aim[i]] = deep[x] + 1;
				q.push(aim[i]);
				if(aim[i] == T)	return true;	
			}
	}
	return false;
}

int Dinic(int x,int f)
{
	if(x == T)	return f;
	int temp = f;
	for(int i = head[x]; i; i = next[i])
		if(flow[i] && temp && deep[aim[i]] == deep[x] + 1) {
			int away = Dinic(aim[i],min(flow[i],temp));
			if(!away)	deep[aim[i]] = 0;
			flow[i] -= away;
			flow[i^1] += away;
			temp -= away;
		}
	return f - temp;
}

int main()
{
	cin >> points >> edges >> S >> T;
	S <<= 1,T = (T << 1|1);
	for(int x,i = 1; i <= points; ++i) {
		scanf("%d",&x);
		Add(i << 1,i << 1|1,x);
		Add(i << 1|1,i << 1,0);
	}
	for(int x,y,i = 1; i <= edges; ++i) {
		scanf("%d%d",&x,&y);
		Add(x << 1|1,y << 1,INF);
		Add(y << 1,x << 1|1,0);
		Add(y << 1|1,x << 1,INF);
		Add(x << 1,y << 1|1,0);
	}
	int max_flow = 0;
	while(BFS())
		max_flow += Dinic(S,INF);
	cout << max_flow << endl;
	return 0;
}