POJ3281 Dining

 经典的网络流建模题，开始其实就想到food连牛连drink，但没想到怎么控制一头牛只吃一份food和一份drink，后来换了一种想法是一头牛连一头左牛，一头右牛，分别连饮料，食物，这样可以做到把一头牛吃的食物和drink分开来，但还是不能解决一头牛各吃一份，最后想起开始的思路，发现一头牛可以拆成左牛右牛控制流量啊，bingo。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define maxn 1000
struct edge {int to,cap,rev;};
int level [maxn];
vector <edge> G[maxn];
int iter[maxn];
void add_edge (int from,int to,int cap){
    G[from].push_back ((edge){to,cap,static_cast<int>(G[to].size())});
    G[to].push_back ((edge){from,0,static_cast<int>(G[from].size()-1)});
}

void bfs (int s){
    memset (level,-1,sizeof (level));
    queue<int> que;
    level[s] = 0;
    que.push (s);
    while (!que.empty()){
        int v=que.front ();
        que.pop ();
        int i;
        for (i=0;i<G[v].size();i++){
            edge &e = G[v][i];
            if (e.cap>0 && level[e.to]<0){
                level[e.to]=level[v]+1;
                que.push(e.to);
            }
        }
    }
}
int dfs (int v,int t,int f){
    if (v==t) return f;
    for (int &i=iter[v];i<G[v].size();i++){
        edge &e =G[v][i];
        if (e.cap>0 && level[v]<level[e.to]){
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0){
                e.cap-=d;
                G[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return 0;
}

int maxflow (int s,int t){
    int flow = 0;
    for (;;){
        bfs (s);
        if (level[t]<0) return flow;
        memset (iter, 0,sizeof(iter));
        int f;
        while ((f=dfs(s,t,inf))>0){
            flow += f;
        }
    }
}
int main (){
    int n,f,d;
    int i;
    while (scanf ("%d %d %d",&n,&f,&d)!=EOF){
        int food = 1;
        int cow = f+1;
        int cow2 = n+d+f+1;
        int t = cow2+n;
        int dnk = n+f+1;
        for (i=1;i<=n;i++){
            int F,D;
            add_edge(cow+i-1, cow2+i-1, 1);
            scanf ("%d %d",&F,&D);
            for (int j=1;j<=F;j++){
                int v;
                scanf ("%d",&v);
                add_edge (v,cow+i-1,1);
            }
            for (int j=1;j<=D;j++){
                int v;
                scanf("%d",&v);
                add_edge (cow2+i-1,dnk+v-1,1);
            }
        }
        for (i=1;i<=f;i++){
            add_edge (0,food+i-1,1);
        }
        for (i=1;i<=d;i++){
            add_edge (dnk+i-1,t,1);
        }
        printf("%d\n", maxflow(0,t));
    }
    return 0;
}