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
本題是要求圖中的最大完全子圖（最大團(tuán)）中頂點(diǎn)的個(gè)數(shù)。由于原圖的補(bǔ)圖是一個(gè)二分圖，其最大完全數(shù)等價(jià)于其補(bǔ)圖的最大獨(dú)立集中元素的個(gè)數(shù)，于是可以根據(jù)二分圖的性質(zhì)求出這個(gè)最大獨(dú)立集。而普通圖的最大團(tuán)則是一個(gè)NP問題。
定理：二分圖最大獨(dú)立集=頂點(diǎn)數(shù)-二分圖最大匹配
最大完全數(shù)：圖中最大完全子圖的頂點(diǎn)個(gè)數(shù)。
獨(dú)立集：圖中任意兩個(gè)頂點(diǎn)都不相連的頂點(diǎn)集合。
#include <iostream>
using namespace std;
const int MAXN = 201;
bool visit[MAXN];
int n,m,k,mark[MAXN];
bool map[MAXN][MAXN];
bool dfs(int u)
{
int i;
for(i=1;i<=m;i++)
if(map[u][i] && !visit[i]){
visit[i]=true;
if(mark[i]==-1 || dfs(mark[i])){
mark[i]=u;
return true;
}
}
return false;
}
int hungary(){
int i,ans=0;
for(i=1;i<=n;i++){
memset(visit,false,sizeof(visit));
if(dfs(i)) ans++;
}
return ans;
}
int main(){
int i,j,x,y,id=1;
while(scanf("%d %d %d",&n,&m,&k),n||m||k){
for(i=1;i<=n;i++) for(j=1;j<=m;j++) map[i][j]=true;
while(k--){
scanf("%d %d",&x,&y);
map[x][y]=false;
}
memset(mark,-1,sizeof(mark));
printf("Case %d: %d\n",id++,n+m-hungary());
}
return 0;
}