hdu 2458 Kindergarten

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2458

Kindergarten

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1286    Accepted Submission(s): 679

Problem Description

In a kindergarten, there are a lot of kids. All girls of the kids know each other and all boys also know each other. In addition to that, some girls and boys know each other. Now the teachers want to pick some kids to play a game, which need that all players know each other. You are to help to find maximum number of kids the teacher can pick.

 

Input

The input consists of multiple test cases. Each test case starts with a line containing three integers
G, B (1 ≤ G, B ≤ 200) and M (0 ≤ M ≤ G × B), which is the number of girls, the number of boys and
the number of pairs of girl and boy who know each other, respectively.
Each of the following M lines contains two integers X and Y (1 ≤ X≤ G,1 ≤ Y ≤ B), which indicates that girl X and boy Y know each other.
The girls are numbered from 1 to G and the boys are numbered from 1 to B.

The last test case is followed by a line containing three zeros.

 

Output

For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the maximum number of kids the teacher can pick.

 

Sample Input

2 3 3

1 1

1 2

2 3

2 3 5

1 1

1 2

2 1

2 2

2 3

0 0 0

 

Sample Output

Case 1: 3

Case 2: 4

 

题目大意：有一堆男孩和女孩，男孩和男孩之间，女孩和女孩之间互相认识，给出一堆男孩女孩之间认识的关系，

　　　　　问最多能邀请几个人去party，（邀请人之间互相直接认识）。

解题思路：把不认识的标记为1,认识的标记为0,那么这个结果就是该二分图的最大独立点集.最大匹配数=最小点覆盖=n-最大点独立集

AC代码：

#include<stdio.h> 
#include<string.h>  
#define N 210  
int map[N][N],v[N],link[N];  
int n,m;  
int dfs(int k)  
{  
    int i;  
    for(i=1;i<=m;i++)  
    {  
        if(map[k][i]&&!v[i])  
        {  
            v[i]=1;  
            if(link[i]==-1||dfs(link[i]))  
            {  
                link[i]=k;  
                return 1;  
            }  
        }  
    }  
    return 0;  
}  
int main()  
{  
    int i,j,ans,a,b,t,tt=0;
    while(~scanf("%d%d%d",&n,&m,&t))  
    {  
        if(!n&&!m&&!t)  
            break;  
        memset(map,1,sizeof(map));  
        while(t--)  
        {  
            scanf("%d%d",&a,&b);  
            map[a][b]=0;  
        }  
        ans=0;  
        memset(link,-1,sizeof(link));  
        for(i=1;i<=n;i++)  
        {  
            memset(v,0,sizeof(v));  
            if(dfs(i))  
                ans++;  
        }  
        printf("Case %d: %d
",++tt,n+m-ans);  
    }  
    return 0;  
}