POJ 3281 Dining（最大流）

Dining

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6586		Accepted: 3015

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

 

本题能够想到用最大流做，那真的是太绝了。建模的方法很妙！

题意就是有N头牛，F个食物，D个饮料。

N头牛每头牛有一定的喜好，只喜欢几个食物和饮料。

每个食物和饮料只能给一头牛。一头牛只能得到一个食物和饮料。

而且一头牛必须同时获得一个食物和一个饮料才能满足。问至多有多少头牛可以获得满足。

最初相当的是二分匹配。但是明显不行，因为要分配两个东西，两个东西还要同时满足。

最大流建图是把食物和饮料放在两端。一头牛拆分成两个点，两点之间的容量为1.喜欢的食物和饮料跟牛建条边，容量为1.

加个源点和汇点。源点与食物、饮料和汇点的边容量都是1，表示每种食物和饮料只有一个。

这样话完全是最大流问题了，。

/*
POJ 3281 最大流
//源点-->food-->牛(左)-->牛(右)-->drink-->汇点
//精髓就在这里，牛拆点，确保一头牛就选一套food和drink的搭配

*/

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>
using namespace std;

//****************************************************
//最大流模板
//初始化：g[][],start,end
//******************************************************
const int MAXN=500;
const int INF=0x3fffffff;
int g[MAXN][MAXN];//存边的容量，没有边的初始化为0
int path[MAXN],flow[MAXN],start,end;
int n;//点的个数，编号0-n.n包括了源点和汇点。

queue<int>q;
int bfs()
{
    int i,t;
    while(!q.empty())q.pop();//把清空队列
    memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1
    path[start]=0;
    flow[start]=INF;//源点可以有无穷的流流进
    q.push(start);
    while(!q.empty())
    {
        t=q.front();
        q.pop();
        if(t==end)break;
        //枚举所有的点，如果点的编号起始点有变化可以改这里
        for(i=0;i<=n;i++)
        {
            if(i!=start&&path[i]==-1&&g[t][i])
            {
                flow[i]=flow[t]<g[t][i]?flow[t]:g[t][i];
                q.push(i);
                path[i]=t;
            }
        }
    }
    if(path[end]==-1)return -1;//即找不到汇点上去了。找不到增广路径了
    return flow[end];
}
int Edmonds_Karp()
{
    int max_flow=0;
    int step,now,pre;
    while((step=bfs())!=-1)
    {
        max_flow+=step;
        now=end;
        while(now!=start)
        {
            pre=path[now];
            g[pre][now]-=step;
            g[now][pre]+=step;
            now=pre;
        }
    }
    return max_flow;
}
int main()
{
    int N,F,D;
    while(scanf("%d%d%d",&N,&F,&D)!=EOF)
    {
        memset(g,0,sizeof(g));
        n=F+D+2*N+1;
        start=0;
        end=n;
        for(int i=1;i<=F;i++)g[0][i]=1;
        for(int i=F+2*N+1;i<=F+2*N+D;i++)g[i][n]=1;
        for(int i=1;i<=N;i++)g[F+2*i-1][F+2*i]=1;
        int k1,k2;
        int u;
        for(int i=1;i<=N;i++)
        {
            scanf("%d%d",&k1,&k2);
            while(k1--)
            {
                scanf("%d",&u);
                g[u][F+2*i-1]=1;
            }
            while(k2--)
            {
                scanf("%d",&u);
                g[F+2*i][F+2*N+u]=1;
            }
        }
        printf("%d\n",Edmonds_Karp());
    }
    return 0;
}