USACO Section 4.2: The Perfect Stall

这题关键就在将题转换成最大流模板题。首先有一个原始点，N个cow个点， M个barn点和一个终点，原始点到cow点和barn点到终点的流都为1，而cow对应的barn就是cow点到对应barn点的流，为1.这样题目就转换成了原始点到终点的最大流问题

 /*
 ID: yingzho1
 LANG: C++
 TASK: stall4
 */
 #include <iostream>
 #include <fstream>
 #include <string>
 #include <map>
 #include <vector>
 #include <set>
 #include <algorithm>
 #include <stdio.h>
 #include <queue>
 #include <cstring>
 #include <cmath>
 #include <list>
 #include <cstdio>
 #include <cstdlib>
 #include <limits>
 #include <stack>

 using namespace std;

 ifstream fin("stall4.in");
 ofstream fout("stall4.out");

 ;
 ;

 int N, M;
 *MAX][*MAX], f[*MAX][*MAX], pre[*MAX], inc[*MAX];

 bool bfs(int s, int d) {
     queue<int> que;
     ; i <= N+M+; i++) pre[i] = -;
     que.push(s);
     inc[s] = INF;
     while (!que.empty()) {
         int u = que.front();
         que.pop();
         ; i <= N+M+; i++) {
              && f[u][i] < g[u][i]) {
                 inc[i] = min(inc[u], g[u][i]-f[u][i]);
                 pre[i] = u;
                 if (i == d) return true;
                 que.push(i);
             }
         }
     }
     return false;
 }

 int edmond_karp(int s, int d) {
     ;
     while (bfs(s, d)) {
         maxflow += inc[d];
         for (int i = d; i != s; i = pre[i]) {
             f[pre[i]][i] += inc[d];
             f[i][pre[i]] -= inc[d];
         }
     }
     return maxflow;
 }

 int main()
 {
     fin >> N >> M;
     int s, d;
     ; i <= N; i++) {
         g[][+i] = ;
         fin >> s;
         ; j < s; j++) {
             fin >> d;
             g[+i][+N+d] = ;
         }
     }
     ; i <= N+M+; i++) g[i][N+M+] = ;
     fout << edmond_karp(, N+M+) << endl;

     ;
 }