Kindergarten（二分图）

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 31 11 22 32 3 51 11 22 12 22 30 0 0 

 

Sample Output

 Case 1: 3Case 2: 4 

解题思路

建图时互相不认识的为一个链接，就是求不认识的最小覆盖数即完美匹配数，从n + m中减去它即可。

AC代码

#include <cstdio>#include <cstring>#define N 205int map[N][N], match[N], vis[N];int n, m, k;int find(int i){    for(int j = 1; j <= m; j++)    {        if(!vis[j] && !map[i][j])        {            vis[j] = 1;            if(match[j] == -1 || find(match[j]))            {                match[j] = i;                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 += find(i);    }    return ret;}int main(){    int x, y;    int cas = 1;    while(scanf("%d%d%d", &n, &m, &k), n + m + k)    {        memset(map, 0, sizeof(map));        for(int i = 0; i < k; i++)        {            scanf("%d%d", &x, &y);            map[x][y] = 1;        }        printf("Case %d: %d\n", cas++, n + m - hungary());    }    return 0;}