poj3281网络流之最大流

加一个源点和汇点，把每头牛拆成两个点，不拆点的话可能会出现多对食物与饮料被一个牛享用的情况，拆点后流量为1，不能同时通过了

然后用最大流处理，每个链接边都是1

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<cassert>
#include<iomanip>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define pi acos(-1)
#define ll long long
#define mod 1000000007
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

const double g=10.0,eps=1e-9;
const int N=400+5,maxn=16,inf=999999;

int c[N][N],pre[N];
bool vis[N];
int s,t,n,food,drink;
bool bfs()
{
    memset(pre,0,sizeof pre);
    memset(vis,0,sizeof vis);
    vis[s]=1;
    queue<int>q;
    q.push(s);
    while(!q.empty()){
        int k=q.front();
        if(k==t)return 1;
        q.pop();
        for(int i=0;i<=n+n+food+drink+1;i++)
        {
            if(!vis[i]&&c[k][i])
            {
                vis[i]=1;
                q.push(i);
                pre[i]=k;
            }
        }
    }
    return 0;
}
int max_flow()
{
    int ans=0;
    while(1){
        if(!bfs())return ans;
        int minn=9999999;
        for(int i=t;i!=s;i=pre[i])
            minn=min(minn,c[pre[i]][i]);
        for(int i=t;i!=s;i=pre[i])
        {
            c[pre[i]][i]-=minn;
            c[i][pre[i]]+=minn;
        }
        ans+=minn;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>food>>drink;
    memset(c,0,sizeof c);
    for(int i=1;i<=n;i++)
    {
        int a,b,k;
        cin>>a>>b;
        while(a--){
            cin>>k;
            c[k][i+food]=1;
        }
        while(b--){
            cin>>k;
            c[i+n+food][k+n+n+food]=1;
        }
    }
    s=0,t=n+n+food+drink+1;
    for(int i=1;i<=n;i++)c[food+i][food+n+i]=1;
    for(int i=1;i<=food;i++)//源点为0，汇点为n+food+drink+1
        c[0][i]=1;
    for(int i=1;i<=drink;i++)
        c[n+n+food+i][n+n+food+drink+1]=1;
    cout<<max_flow()<<endl;
    return 0;
}