poj 3281 Dining

Dining

 POJ - 3281 

题目大意：

n头牛，f种食物，d种饮料， 每头牛有喜欢的食物、饮料， 每头牛只能分配一种食物和一种饮料 问最后最多能满足多少头牛

/*
    每头牛只能选择一种食物和一种饮料，拆点可以进行容量限制 
*/
#include<iostream>
#include<cstdio>
#include<queue>
#define maxn 501
using namespace std;
int n,f,d,num=1,lev[maxn],head[maxn],cur[maxn],t,ans;
struct node{
    int to,pre,cap;
}e[maxn*maxn*2];
void Insert(int from,int to,int v){
    e[++num].to=to;
    e[num].cap=v;
    e[num].pre=head[from];
    head[from]=num;
}
bool bfs(){
    queue<int>q;
    for(int i=0;i<=t;i++)cur[i]=head[i],lev[i]=-1;
    q.push(0);
    lev[0]=0;
    while(!q.empty()){
        int now=q.front();q.pop();
        for(int i=head[now];i;i=e[i].pre){
            int to=e[i].to;
            if(lev[to]==-1&&e[i].cap>0){
                lev[to]=lev[now]+1;
                q.push(to);
                if(to==t)return 1;
            }
        }
    }
    return 0;
}
int dinic(int now,int flow){
    if(now==t)return flow;
    int rest=0,delta;
    for(int &i=cur[now];i;i=e[i].pre){
        int to=e[i].to;
        if(lev[to]==lev[now]+1&&e[i].cap>0){
            delta=dinic(to,min(e[i].cap,flow-rest));
            if(delta){
                e[i].cap-=delta;e[i^1].cap+=delta;
                rest+=delta;
                if(rest==flow)break;
            }
        }
    }
    if(rest!=flow)lev[now]=-1;
    return rest;
}
int main(){
    freopen("Cola.txt","r",stdin);
    scanf("%d%d%d",&n,&f,&d);
    t=f+2*n+d+1;
    for(int i=1;i<=n;i++){
        Insert(i+f,i+n+f,1);
        Insert(i+n+f,i+f,0);
    }
    for(int i=1;i<=f;i++){
        Insert(0,i,1);
        Insert(i,0,0);
    }
    int a,b,x,y,z;
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a,&b);
        for(int j=1;j<=a;j++){//食物向牛连边 
            scanf("%d",&x);
            Insert(x,i+f,1);
            Insert(i+f,x,0);
        }
        for(int j=1;j<=b;j++){//牛向饮料连边 
            scanf("%d",&x);
            Insert(i+f+n,x+2*n+f,1);
            Insert(x+2*n+f,i+f+n,0);
        }
    }
    for(int i=1;i<=d;i++){//饮料向汇点连边 
        Insert(i+f+2*n,t,1);
        Insert(t,i+f+2*n,0);
    }
    while(bfs())
        ans+=dinic(0,9999999);
    printf("%d",ans);
}