题意：有一个犯罪集团要贩卖大规模杀伤武器，从s城运输到t城，现在你是一个特殊部门的长官，可以在城市中布置眼线，但是布施眼线需要花钱，现在问至少要花费多少能使得你及时阻止他们的运输。
题解：裸的最小割模型，最小割就是最大流，我们把点拆成2个点，然后将原点与拆点建边，流量为在城市建立眼线的费用，然后拆点为出点，原点为入点，将可以到达的城市之间建流量为无穷的边。
最后求出s 到 t的拆点的最大流 那么就是这个题目的答案了。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb emplace_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define lch(x) tr[x].son[0] 12 #define rch(x) tr[x].son[1] 13 #define max3(a,b,c) max(a,max(b,c)) 14 #define min3(a,b,c) min(a,min(b,c)) 15 typedef pair<int,int> pll; 16 const int inf = 0x3f3f3f3f; 17 const LL INF = 0x3f3f3f3f3f3f3f3f; 18 const LL mod = (int)1e9+7; 19 const int N = 500; 20 const int M = N*N*4; 21 int head[N], deep[N], cur[N]; 22 int w[M], to[M], nx[M]; 23 int tot; 24 int n, m, s, t; 25 int u, v, val; 26 void add(int u, int v, int val){ 27 w[tot] = val; to[tot] = v; 28 nx[tot] = head[u]; head[u] = tot++; 29 30 w[tot] = 0; to[tot] = u; 31 nx[tot] = head[v]; head[v] = tot++; 32 } 33 int bfs(int s, int t){ 34 queue<int> q; 35 memset(deep, 0, sizeof(deep)); 36 q.push(s); 37 deep[s] = 1; 38 while(!q.empty()){ 39 int u = q.front(); 40 q.pop(); 41 for(int i = head[u]; ~i; i = nx[i]){ 42 if(w[i] > 0 && deep[to[i]] == 0){ 43 deep[to[i]] = deep[u] + 1; 44 q.push(to[i]); 45 } 46 } 47 } 48 return deep[t] > 0; 49 } 50 int Dfs(int u, int t, int flow){ 51 if(u == t) return flow; 52 for(int &i = cur[u]; ~i; i = nx[i]){ 53 if(deep[u]+1 == deep[to[i]] && w[i] > 0){ 54 int di = Dfs(to[i], t, min(w[i], flow)); 55 if(di > 0){ 56 w[i] -= di, w[i^1] += di; 57 return di; 58 } 59 } 60 } 61 return 0; 62 } 63 64 int Dinic(int s, int t){ 65 int ans = 0, tmp; 66 while(bfs(s, t)){ 67 for(int i = 1; i <= 2*n+2; i++) cur[i] = head[i]; 68 while(tmp = Dfs(s, t, inf)) ans += tmp; 69 } 70 return ans; 71 } 72 void init(){ 73 memset(head, -1, sizeof(head)); 74 tot = 0; 75 } 76 int main(){ 77 while(~scanf("%d%d", &n, &m)){ 78 init(); 79 int ss = n*2+1, tt = ss+1; 80 s = ss, t = tt; 81 scanf("%d%d", &ss, &tt); 82 add(s, ss, inf); 83 add(tt+n, t, inf); 84 for(int i = 1; i <= n; i++){ 85 scanf("%d", &val); 86 add(i,i+n,val); 87 } 88 for(int i = 1; i <= m; i++){ 89 scanf("%d%d",&u,&v); 90 add(u+n,v,inf); 91 add(v+n,u,inf); 92 } 93 printf("%d ",Dinic(s,t)); 94 } 95 return 0; 96 }