二分图的最大团

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 ： 所有的男生都认识，所有的女生都认识，让你寻找一个最大的一个团，使得其中的所有人之间彼此都认识思路分析 ： 题目是让寻找一个二分图的最大团，最大团 = 补图的最大独立集 = 顶点数 - 补图的最小顶点覆盖数 = 顶点数 - 补图的二分图匹配数代码示例 ：

#define ll long long
const int maxn = 1e6+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

bool edge[205][205];
int girl, boy, m;
bool used[205];
int pt[205];

bool dfs(int x){
    for(int i = 1; i <= boy; i++){
        if (edge[x][i] && !used[i]) {
            used[i] = true;
            if (!pt[i] || dfs(pt[i])){
                pt[i] = x;
                return true;
            }
        }
    }
    return false;
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int a, b;
    int kase = 1;
    while(scanf("%d%d%d", &girl, &boy, &m) && girl+boy+m){
        memset(edge, true, sizeof(edge));
        for(int i = 1; i <= m; i++){
            scanf("%d%d", &a, &b);
            edge[a][b] = false;
        }
        int cnt = 0;
        memset(pt, 0, sizeof(pt));
        for(int i = 1; i <= girl; i++){
            memset(used, false, sizeof(used));
            if (dfs(i)) cnt++;
        }
        printf("Case %d: %d\n", kase++, girl+boy-cnt);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/ccut-ry/p/8511536.html