Dining （匹配，最大流）

题目链接
题目大意：奶牛喜欢一些种类的食物和饮料，给n种食物，m种饮料，问最多能喂多少奶牛（每种食物和饮料只能使用一次）
思路：把奶牛,食物,饮料都看做点，为了保证路线合法性，奶牛需要在中间，向2边构图。奶牛需要拆点。

emmmm如果奶牛喜欢三样东西应该怎么构图？

#include <bits/stdc++.h>

using namespace std;
const int N = 110*4;
const int M = N*N;
int n,m,s,t,tot;
int head[N];
struct node{
    int u,nxt,cap,to;
}edge[M<<1];
int cur[N],deep[N];
void ae(int u,int v,int w){
    tot++;
    edge[tot].nxt = head[u];
    edge[tot].to = v;
    edge[tot].u = u;
    edge[tot].cap = w;
    head[u] = tot;
}
bool bfs(){
    memset(deep,-1,sizeof(deep));
    queue<int>q;
    q.push(s);
    deep[s] = 0;
    while(!q.empty()){
        int u = q.front();
        for(int i = head[u]; ~i;i = edge[i].nxt){
            int v = edge[i].to;
            if(deep[v]==-1&&edge[i].cap>0){
                q.push(v);
                deep[v] = deep[u]+1;
                if(v==t) return 1;
            }
        }
        q.pop();
    }
    return 0;
}
int dfs(int u,int f){
    int flow = 0,d;
    if(u==t||f==0) return f;
    for(int &i = cur[u]; ~i;i = edge[i].nxt){
        int v = edge[i].to;
        if(deep[v]>deep[u]&&edge[i].cap>0&&(d=dfs(v,min(f,edge[i].cap)))){
            edge[i].cap -= d;
            edge[i^1].cap += d;
            f -= d;
            flow += d;
            if(!f) break;
        }
    }
    if(flow==0) deep[u] = -1;
    return flow;
}
int dinic(){
    int ans = 0;
    while(bfs()){
        memcpy(cur, head, sizeof(head));
        ans += dfs(s,1e9);
    }
    return ans;
}
int main()
{
    //freopen("a.txt","r",stdin);
    ios::sync_with_stdio(0);
    int n1,n2,n3;
    cin>>n2>>n1>>n3;
    s = 0;
    t = n1+n2*2+n3+1;  
    memset(head,-1,sizeof(head));
    tot = -1;
    int i,j;
    for(i = 1;i <= n1;i ++){
        ae(0,i,1); //s连食物
        ae(i,0,0);
    }
    for(i = n1+n2*2+1;i <= n1+n2*2+n3;i ++){
        ae(i,t,1);  //饮料连t
        ae(t,i,0);
    }
    for(i = 1;i <= n2;i ++){
        ae(n1+i,n1+i+n2,1);  //奶牛拆点
        ae(n1+i+n2,n1+i,0);
        int num1,num2;
        cin>>num1>>num2;
        for(j = 1;j <= num1;j ++){
            int x;
            cin>>x;
            ae(x,n1+i,1);  //奶牛连食物
            ae(n1+i,x,0);
        }
        for(j = 1;j <= num2;j ++){
            int x;
            cin>>x;
            ae(n1+i+n2,n1+n2*2+x,1);  //奶牛连饮料
            ae(n1+n2*2+x,n1+i+n2,0);
        }
    }
    cout<<dinic();
    return 0;
}