POJ-3692Kindergarten,求最大独立集！

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K

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

题意：在所有男孩都认识彼此与所有女孩都认识彼此的情况下，求一个人数最多的集合使得集合中任意两个人互相认识。

思路：顶点数=最大独立数+最小点覆盖数（最大匹配数）；这里我们建立补图，补图有边的说明是不认识的，那么用最少的点（补图的最大二分匹配）将这些边去掉剩下的就全部互相认识了；

const int N=200+10;
int G,B,m,g[N][N],linked[N],v[N];
int dfs(int u)
{
    for(int i=1; i<=B; i++)
        if(!v[i]&&g[u][i])
        {
            v[i]=1;
            if(linked[i]==-1||dfs(linked[i]))
            {
                linked[i]=u;
                return 1;
            }
        }
    return 0;
}
void hungary()
{
    int ans=0;
    memset(linked,-1,sizeof(linked));
    for(int i=1; i<=G; i++)
    {
        memset(v,0,sizeof(v));
        if(dfs(i)) ans++;
    }
    printf("%d\n",G+B-ans);
}
int main()
{
    int t1=1;
    while(~scanf("%d%d%d",&G,&B,&m))
    {
        if(G==m&&G==B&&B==0) break;
        int x,y;
        memset(g,-1,sizeof(g));
        while(m--)
        {
            scanf("%d%d",&x,&y);
            g[x][y]=0;
        }
        printf("Case %d: ",t1++);
        hungary();
    }
    return 0;
}

关键在于建立补图然后求出最大二分匹配，只要构图和建模清晰了，剩下的就是裸模板了。