POJ 3692 Kindergarten（最大团问题）

题目链接：
http://poj.org/problem?id=3692

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

 /*
 问题
 输入女孩和男孩的个数和部分女孩和男孩的认识关系
 计算并输出最多有多少个男孩和女孩是相互认识的

 解题思路
 属于图论中的最大团问题。了解了完全子图，意即该子图中任意两点相互连接，那么最大完全子图就是求解
 最大完全子图的顶点数，也就是俗语说的最大团问题了。
 再来说说如何求解最大完全子图顶点数
 有一个定理：最大团=原图补图的最大独立集=顶点数-原图补图最大匹配数

 那怎么什么是补图呢，其实就是原图的完全相反状态，1就是0,0就是1
 所以和求原图的最大独立集算法中只需将注释出改一个相反即可。
 */
 #include<cstdio>
 #include<cstring>

 int g,b,m;
 int e[][],cx[],cy[],bk[];
 int maxmatch();
 int path(int u);
 int main()
 {
     int t1,t2,t=,i;
     while(scanf("%d%d%d",&g,&b,&m) ==  && g+b+m != ){
         memset(e,,sizeof(int)**);
         for(i=;i<=m;i++){
             scanf("%d%d",&t1,&t2);
             e[t1][t2]=;
         }
         printf("Case %d: %d\n",t++,g+b-maxmatch());
     }
     return ;
 }
 int maxmatch()
 {
     int i,ans=;
     memset(cx,-,sizeof(cx));
     memset(cy,-,sizeof(cy));
     for(i=;i<=g;i++){
         memset(bk,,sizeof(bk));
         ans += path(i);
     }
     return ans;
 }
 int path(int u)
 {
     int i;
     for(i=;i<=b;i++){
         if(!e[u][i] && !bk[i]){//求补图的最大独立集，所以只需将原来的e[u][i]变为!e[u][i]即可
             bk[i]=;
             if(cy[i] == - || path(cy[i])){
                 cy[i]=u;
                 cx[u]=i;
                 return ;
             }
         }
     }
     return ;
 }