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");

const int MAX = 210;
const int INF = 2000000000;

int N, M;
int g[2*MAX][2*MAX], f[2*MAX][2*MAX], pre[2*MAX], inc[2*MAX];

bool bfs(int s, int d) {
queue<int> que;
for (int i = 1; i <= N+M+2; i++) pre[i] = -1;
que.push(s);
inc[s] = INF;
while (!que.empty()) {
int u = que.front();
que.pop();
for (int i = 1; i <= N+M+2; i++) {
if (pre[i] == -1 && 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) {
int maxflow = 0;
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;
for (int i = 1; i <= N; i++) {
g[1][1+i] = 1;
fin >> s;
for (int j = 0; j < s; j++) {
fin >> d;
g[1+i][1+N+d] = 1;
}
}
for (int i = N+2; i <= N+M+1; i++) g[i][N+M+2] = 1;
fout << edmond_karp(1, N+M+2) << endl;

return 0;
}