POJ

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 F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i 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

Sample Input

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

Sample Output

3

Hint

One way to satisfy three cows is:
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only three kinds of food or drink. Other test data sets are more challenging, of course.

题解：

拆点（cow,cow`），建立超级源点和超级汇点，超级源点连接所有的food，food连接cow，cow连接cow`，cow`连接drink，所有drink又连接超级汇点。然后就是基础的DINIC跑一遍。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 410;//F:100 + D:100 + N:100*2

struct Edge{
	int to,value,rev;
	Edge(){}
	Edge(int a,int b,int c):to(a),value(b),rev(c){}
};

vector<Edge> E[MAXN];

inline void Add(int from,int to,int value){
	E[from].push_back(Edge(to,value,E[to].size()));
	E[to].push_back(Edge(from,0,E[from].size()-1));
}

int deep[MAXN];
int iter[MAXN];

bool BFS(int from,int to){
	memset(deep,-1,sizeof deep);
	deep[from] = 0;
	queue<int> Q;
	Q.push(from);
	while(!Q.empty()){
		int t = Q.front();
		Q.pop();
		for(int i=0 ; i<E[t].size() ; ++i){
			Edge& e = E[t][i];
			if(e.value > 0 && deep[e.to] == -1){
				deep[e.to] = deep[t] + 1;
				Q.push(e.to);
			}
		}
	}
	return deep[to] != -1;
}

int DFS(int from,int to,int flow){
	if(from == to || flow == 0)return flow;
	
	for(int& i=iter[from] ; i<E[from].size() ; ++i){
		Edge& e = E[from][i];
		if(e.value > 0 && deep[e.to] == deep[from]+1){
			int nowflow = DFS(e.to,to,min(flow,e.value));
			if(nowflow > 0){
				e.value -= nowflow;
				E[e.to][e.rev].value += nowflow;
				return nowflow;
			}
		}
	}
	return 0;
}


int Dinic(int from,int to){
	int sumflow = 0;
	while(BFS(from,to)){
		memset(iter,0,sizeof iter);
		int mid;
		while((mid=DFS(from,to,INF)) > 0)sumflow += mid;
	}
	return sumflow;
}

int main(){
	
	int N,F,D;
	int Fi,Di;
	while(scanf("%d %d %d",&N,&F,&D)!=EOF){
		//固定超级源点为0，超级汇点为MAXN-1 
		for(int i=1 ; i<=F ; i++)Add(0,i+200,1); 
		for(int i=1 ; i<=D ; i++)Add(i+300,MAXN-1,1);
		for(int _=1 ; _<=N ; ++_){
			scanf("%d %d",&Fi,&Di);
			Add(_,_+100,1);//拆点 
			for(int i=0 ; i<Fi ; ++i){
				int t;
				scanf("%d",&t);
				Add(t+200,_,1);
			}
			for(int i=0 ; i<Di ; ++i){
				int t;
				scanf("%d",&t);
				Add(_+100,t+300,1);
			}
		}
		printf("%d
",Dinic(0,MAXN-1));
		for(int i=0 ; i<MAXN ; ++i)E[i].clear();
	}
	
	return 0;
}