Poj 1273 Drainage Ditches(最大流 Edmonds-Karp )

题目链接：poj1273 Drainage Ditches

呜呜，今天自学网络流，看了EK算法，学的晕晕的，留个简单模板题来作纪念、、、

 #include<cstdio>
 #include<vector>
 #include<queue>
 #include<cstring>
 #include<set>
 #include<algorithm>
 #define CLR(a,b) memset((a),(b),sizeof((a)))
 using namespace std;

 const int N = ;
 const int inf = 0x3f3f3f3f;
 int n, m;
 int g[N][N];//i到j的容量
 int pre[N];//i的前驱节点
 bool vis[N];
 int flow[N][N];//从i到j的流量
 int a[N];//从源点到i节点的最小残留量

 int Edmonds_Karp(int st, int ed){
     int u, v;
     int f = ;//最大流
     queue<int>q;//bfs找增广路
     while(){
         CLR(vis, ); CLR(a, );//每找一次记得初始化
         q.push(st);
         vis[st] = ;
         a[st] = inf;
         while(!q.empty()){
             u = q.front(); q.pop();
             for(v = ; v <= n; ++v){
                 if(!vis[v] && g[u][v] > flow[u][v]){
                     q.push(v);
                     pre[v] = u;
                     vis[v] = ;
                     a[v] = min(a[u], g[u][v] - flow[u][v]);
                 }
             }
         }
         if(a[ed] == ) break;//当前已经是最大流
         f += a[ed];
         for(u = ed; u != st; u = pre[u]){
             flow[pre[u]][u] += a[ed];
             flow[u][pre[u]] -= a[ed];
         }

     }
     return f;
 }
 int main(){
     int i, j, k, a, b, c;
     while(scanf("%d %d", &m, &n)==){
         CLR(g, ); CLR(flow, );
         for(i = ; i < m; ++i){
             scanf("%d %d %d", &a, &b, &c);
             g[a][b] += c;//可能有重边
         }
         int max_flow = Edmonds_Karp(, n);
         printf("%d\n", max_flow);
     }
     return ;
 }