POJ3692 Kindergarten

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6882		Accepted: 3402

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

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题目的意思是给出n个男的m个女的，男的互相认识，女的互相认识，和k组关系，问能选出多少个人两两互相认识

思路：求最大团，最大团=补图的最大独立集=点数-补图的最大匹配

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

const int MAXN=1000;
int uN,vN;  //u,v数目
int g[MAXN][MAXN];//编号是0~n-1的
int linker[MAXN];
bool used[MAXN];

bool dfs(int u)
{
    int v;
    for(v=1; v<=vN; v++)
        if(!g[u][v]&&!used[v])
        {
            used[v]=true;
            if(linker[v]==-1||dfs(linker[v]))
            {
                linker[v]=u;
                return true;
            }
        }
    return false;
}
int hungary()
{
    int res=0;
    int u;
    memset(linker,-1,sizeof(linker));
    for(u=1; u<=uN; u++)
    {
        memset(used,0,sizeof(used));
        if(dfs(u))  res++;
    }
    return res;
}

int main()
{
    int m,u,v;
    int q=1;
    while(~scanf("%d%d%d",&uN,&vN,&m)&&(uN||vN||m))
    {
        memset(g,0,sizeof g);
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            g[u][v]=1;
        }
        printf("Case %d: %d\n",q++,uN+vN-hungary());
    }
    return 0;
}