poj —— 1274 The Perfect Stall

　　　　　　　　　　　　　　　　　　　　　　　　　　The Perfect Stall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26274		Accepted: 11672

Description

Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall. 
Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible. 

Input

The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.

Output

For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.

Sample Input

5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2

Sample Output

4

Source

USACO 40

题目描述：农夫约翰刚刚在上周完成了他的新谷仓，配备了所有最新的挤奶技术。 不幸的是，由于工程问题，新谷仓中的所有摊位都不一样。 第一周，农夫约翰随机分配奶牛到摊位，但很快就明白，任何给定的牛只愿意在某些摊位生产牛奶。 最近一周，农民约翰一直收集有关哪些奶牛愿意生产牛奶的数据。 摊位可能只能分配给一只母牛，当然，一只母牛可能只能被分配到一个摊位。考虑到奶牛的偏好，计算可能的奶牛产奶量的最大数量。

 

二分图的最大匹配

代码；

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 505
using namespace std;
bool vis[N];
int n,m,k,x,ans,girl[N],map[N][N];
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
int find(int x)
{
    ;i<=m;i++)
    {
        if(!vis[i]&&map[x][i])
        {
            vis[i]=true;
            ||find(girl[i])){girl[i]=x; ;}
        }
    }
    ;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        ans=;
        memset(map,,sizeof(map));
        memset(girl,-,sizeof(girl));
        ;i<=n;i++)
        {
            k=read();
            ;
        }
        ;i<=n;i++)
        {
            memset(vis,,sizeof(vis));
            if(find(i)) ans++;
        }
        printf("%d\n",ans);
    }
    ;
}