POJ 3692 Kindergarten (二分图 最大团)

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5660		Accepted: 2756

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

field=source&key=2008+Asia+Hefei+Regional+Contest+Online+by+USTC">2008 Asia Hefei Regional Contest Online by USTC



题目链接：

id=3692">http://poj.org/problem?id=3692



题目大意：一些男生和女生，男生们相互都认识。女生们相互都认识，给出男女生的认识关系。要求一个最大的集合，集合中随意两个人都互相认识，求这个最大集合的元素个数



题目分析：二分图最大团问题。依据定理：

二分图最大团＝原图补图的最大独立集

最大独立集＝总点数－最大匹配

用匈牙利算法解出原图补图的最大匹配就可以算出最大团中元素个数 （注：一般图的最大团问题是NP问题）



代码：

#include <cstdio>
#include <cstring>
int const MAX = 205;
int cx[MAX], cy[MAX];
bool vis[MAX], map[MAX][MAX];
int g, b, m;

int DFS(int x)
{
    for(int y = 1; y <= b; y++)
    {
        if(!vis[y] && map[x][y])
        {
            vis[y] = true;
            if(cy[y] == -1 || DFS(cy[y]))
            {
                cy[y] = x;
                cx[x] = y;
                return 1;
            }
        }
    }
    return 0;
}

int MaxMatch()
{
    int ans = 0;
    memset(cx, -1, sizeof(cx));
    memset(cy, -1, sizeof(cy));
    for(int i = 1; i <= g; i++)
    {
        if(cx[i] == -1)
        {
            memset(vis, false, sizeof(vis));
            ans += DFS(i);
        }
    }
    return ans;
}

int main()
{
    int ca = 1;
    while(scanf("%d %d %d", &g, &b, &m) != EOF && (g + b + m))
    {
        memset(map, true, sizeof(map));
        for(int i = 0; i < m; i++)
        {
            int u, v;
            scanf("%d %d", &u, &v);
            map[u][v] = false;
        }
        printf("Case %d: %d\n", ca ++, g + b - MaxMatch());
    }
}