Kindergarten（二分图）

最新推荐文章于 2020-09-08 20:27:05 发布

geguojun

最新推荐文章于 2020-09-08 20:27:05 发布

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分类专栏：二分图文章标签：二分图

本文链接：https://blog.youkuaiyun.com/geguojun/article/details/38611565

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本文深入探讨了信息技术领域的多个细分技术领域，包括前端开发、后端开发、移动开发、游戏开发、大数据开发等，并详细介绍了各领域的核心技术和应用案例。

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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

Sample Output

 Case 1: 3
Case 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 205
int 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;
}