描述

传送门：poj-3281

Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

Farmer John has cooked $F\ (1\ ≤\ F\ ≤\ 100)$ types of foods and prepared $D\ (1\ ≤\ D\ ≤\ 100)$ types of drinks. Each of his $N\ (1\ ≤\ N\ ≤\ 100)$ cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

Input

Line 1: Three space-separated integers: N, F, and D
Lines 2…N+1: Each line i starts with a two integers Fi and Di, the number of dishes that cow i likes and the number of drinks that cow i likes. The next Fi integers denote the dishes that cow i will eat, and the Di integers following that denote the drinks that cow i will drink.

Output

Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

Examples

intput

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

output
1
3

大致题意

农夫为他的 $N\ (1\ ≤\ N\ ≤\ 100)$ 牛准备了 $F\ (1\ ≤\ F\ ≤\ 100)$ 种食物和 $D\ (1\ ≤\ D ≤\ 100)$ 种饮料。每头牛都有各自喜欢的食物和饮料，而每种食物或饮料只能分配给一头牛。最多能有多少头牛可以同时得到喜欢的食物和饮料？

思路

不难发现这道题需要添加一个超级源点（和所有的食物有一条边）和超级汇点（和所有的饮料有一条边），然后每头牛和自己喜欢的食物，饮料连一条边。然后跑一个最大流。

但是这样会有个问题，有可能同一只羊拿了两个饮料和食物，而限制它的方法就是将牛的这个点拆成两个i和i’两个点，这两个点之间连一条容量为1的边，这样一头牛就只能被选一次（选过一次之后i和i’的之间的边就满流了，不能再增加流量）。

由于要把食物，牛，饮料建在一个图中，就把点哈希一下，1-100表示i，101-200表示i’，201-300表示食物，301-400表示饮料，0表示超级源点，401表示超级汇点。

代码

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define repd(i,a,n) for(int i=n;i>=a;i--)
#define CRL(a,x) memset(a,x,sizeof(a))
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=1e4+5;
const int mod=1e9+7;

int Begin[N],d[N],S=0,T=401,cnt=-1,cur[N];//cur就是记录当前点u循环到了哪一条边;

struct node{int to,w,nex;}Edge[100000];
void add(int u,int v,int w){
    Edge[++cnt]=(node){v,w,Begin[u]};
    Begin[u]=cnt;
    Edge[++cnt]=(node){u,0,Begin[v]};
    Begin[v]=cnt;
}

bool bfs(){
    queue<int> Q;
    CRL(d,0);d[S]=1;
    Q.push(S);
    while(!Q.empty()){
        int u=Q.front();Q.pop();
        for(int  i=Begin[u];~i;i=Edge[i].nex){
            int v=Edge[i].to;
            if(!d[v]&&Edge[i].w){
                d[v]=d[u]+1;
                Q.push(v);
            }
            if(d[T]) return true;
        }
    }
    return false;
}

int dfs(int u,int flow){
    if(u==T) return flow;
    for(int &i=cur[u];~i;i=Edge[i].nex){
        int v=Edge[i].to,tflow;
        if(d[v]==d[u]+1&&Edge[i].w){
            tflow=dfs(v,min(Edge[i].w,flow));
            if(tflow){
                Edge[i].w-=tflow;
                Edge[i^1].w+=tflow;
                return tflow;
            }
        }
    }
    return 0;
}

int dinic(){
    int flow=0,tem;
    while(bfs()){
        rep(i,S,T) cur[i]=Begin[i];
        while(tem=dfs(S,inf))
            flow+=tem;
    }
    return flow;
}

int main(){
    int n,F,D,fi,di,x;
    scanf("%d%d%d",&n,&F,&D);
    CRL(Begin,-1);

    rep(i,1,F) add(S,i+200,1);     //超级源点和食物之间连一条边
    rep(i,1,D) add(i+300,T,1);     //饮料和超级汇点之间连一条边

    rep(i,1,n){
        scanf("%d%d",&fi,&di);
        add(i,i+100,1);         //拆点

        rep(j,1,fi){
            scanf("%d",&x);
            add(x+200,i,1);
        }

        rep(j,1,di){
            scanf("%d",&x);
            add(i+100,x+300,1);
        }
    }
    printf("%d\n",dinic());
    return 0;
}