poj 3692 Kindergarten (最大独立集)

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 4903		Accepted: 2387

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

Source

2008 Asia Hefei Regional Contest Online by USTC

 

题意：有n个男孩和m个女孩，男孩间互相认识，女孩间也互相认识，男孩和女孩间部分认识，求一个最大团里面的人都互相认识。

 

思考没转弯..

这道题想了挺久，想直接暴力，因为找不到合适的匹配方法，后来看别人的思路，建图稍稍改一下就行，改成认识的不匹配，匹配不认识的。

然后剔除所有不认识的，剩下的就是互相认识的。

最大独立集= n-最小覆盖集 = n-完美匹配数。

//312K    125MS    C++    878B    2014-06-13 21:55:34
#include<stdio.h>
#include<string.h>
#define N 205       
int g[N][N];
int match[N];
int vis[N];
int n,m,k;
int dfs(int u)
{
    for(int i=1;i<=m;i++)
        if(!vis[i] && !g[u][i]){
            vis[i]=1;
            if(match[i]==-1 || dfs(match[i])){
                match[i]=u;
                return 1;                
            }
        }
    return 0;
}
int hungary()
{
    int ret=0;
    memset(match,-1,sizeof(match));
    for(int i=1;i<=n;i++){
        memset(vis,0,sizeof(vis));
        ret+=dfs(i);
    }
    return ret;
}
int main(void)
{
    int a,b;
    int cas=1;
    while(scanf("%d%d%d",&n,&m,&k),(n+m+k))
    {
        memset(g,0,sizeof(g));
        for(int i=0;i<k;i++){
            scanf("%d%d",&a,&b);
            g[a][b]=1;
        }
        printf("Case %d: %d\n",cas++,n+m-hungary());
    }
    return 0;
}