网络流 dinic

 1 struct Edge {
 2     int v, w;
 3     int next;
 4 }edge[250*250];
 5 int head[50], tot;
 6 void addedge(int u, int v, int w) {
 7     edge[tot].v = v;
 8     edge[tot].w = w;
 9     edge[tot].next = head[u];
10     head[u] = tot++;
11 }
12 int lvl[305];
13 bool bfs(int s, int t) {
14     lvl[s] = 0;
15     queue<int> Q;
16     Q.push(s);
17     while(!Q.empty()) {
18         int tmp = Q.front();
19         Q.pop();
20         if(tmp == t) return true;
21         for(int i = head[tmp]; i+1; i = edge[i].next) {
22             int v = edge[i].v;
23             int w = edge[i].w;
24             if(!lvl[v] && w) {
25                 Q.push(v);
26                 lvl[v] = lvl[tmp] + 1;
27             }
28         }
29     }
30     return false;
31 }
32 int dfs(int now, int t, int F) {
33     if(now == t) return F;
34     int ret = 0, f;
35     for(int i = head[now]; i+1; i = edge[i].next) {
36         int v = edge[i].v;
37         int w = edge[i].w;
38         if(w && lvl[v] == lvl[now] + 1) {
39             f = dfs(v, t, min(F-ret, w));
40             edge[i].w -= f;
41             edge[i^1].w += f;
42             ret += f;
43             if(ret == F) return ret;
44         }
45     }
46     return ret;
47 }
48 int dinic(int s, int t) {
49     int ans = 0;
50     while(bfs(s, t)) ans += dfs(s, t, INF);
51     return ans;
52 }