ＰＯＪ ３２８１ Dining（网络流）

                                                                    Dining

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 16967		Accepted: 7535

Description

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.

Source

USACO 2007 Open Gold

【分析】典型的网络流题，关键是建图。对于每头牛，既要满足得到食物，又要满足得到饮料，所以将源点与所有食物建边，所有饮料与汇点建边，由于每头牛只能占领一种食物和饮料，

所以考虑将牛拆点，ｉ－＞ｉ＋ｎ，这样避免了一头牛占领多种食物和饮料，然后食物到牛，牛到饮料，容量全为１.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 1e3+10;
const int M = 24005;
int n,m,k,f,d;
struct Edge{
    int from,to,cap,flow;
    Edge(int u,int v,int c,int f):from(u),to(v),cap(c),flow(f){}
};
struct Dinic{
    int s,t;
    vector<Edge>edges;
    vector<int> G[N];
    bool vis[N];
    int d[N];
    int cur[N];
    void init(){
       for (int i=0;i<=n+1;i++)
           G[i].clear();
       edges.clear();
    }
    void AddEdge(int from,int to,int cap){
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));
        int mm=edges.size();
        G[from].push_back(mm-2);
        G[to].push_back(mm-1);
    }
    bool BFS(){
        memset(vis,0,sizeof(vis));
        queue<int>q;
        q.push(s);
        d[s]=0;
        vis[s]=1;
        while (!q.empty()){
            int x = q.front();q.pop();
            for (int i = 0;i<G[x].size();i++){
                Edge &e = edges[G[x][i]];
                if (!vis[e.to] && e.cap > e.flow){
                    vis[e.to]=1;
                    d[e.to] = d[x]+1;
                    q.push(e.to);
                }
            }
        }
        return vis[t];
    }

    int DFS(int x,int a){
        if (x==t || a==0)
            return a;
        int flow = 0,f;
        for(int &i=cur[x];i<G[x].size();i++){
            Edge &e = edges[G[x][i]];
            if (d[x]+1 == d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow)))>0){
                e.flow+=f;
                edges[G[x][i]^1].flow-=f;
                flow+=f;
                a-=f;
                if (a==0)
                    break;
            }
        }
        return flow;
    }

    int Maxflow(int s,int t){
        this->s=s;
        this->t=t;
        int flow = 0;
        while (BFS()){
            memset(cur,0,sizeof(cur));
            flow+=DFS(s,inf);
        }
        return flow;
    }
}dc;
int main(){
    scanf("%d%d%d",&n,&f,&d);
    int F,D,u;
    for(int i=1;i<=n;i++){
        scanf("%d%d",&F,&D);
        while(F--){
            scanf("%d",&u);
            dc.AddEdge(2*n+u,i,1);
        }
        while(D--){
            scanf("%d",&u);
            dc.AddEdge(i+n,2*n+f+u,1);
        }
        dc.AddEdge(i,i+n,1);
    }
    for(int i=1;i<=f;i++)dc.AddEdge(0,2*n+i,1);
    for(int i=1;i<=d;i++)dc.AddEdge(2*n+f+i,1000,1);
    printf("%d
",dc.Maxflow(0,1000));
    return 0;
}