POJ 题目3692 Kindergarten（最大独立集）

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5710		Accepted: 2776

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
题目大意：幼儿园有g个女孩和b个男孩，同性之间互相认识，而且男孩和女孩之间有的也互相认识。现在要选出来最多的孩子，他们之间都互相认识。
一道基础的二分图最大独立集问题。

ac代码

#include<stdio.h>
#include<string.h>
int link[220],map[220][220],vis[220];
int x,y,m;
void init()
{
int i,j;
for(i=0;i<=x;i++)
{
for(j=0;j<=y;j++)
map[i][j]=1;
}
}
int dfs(int x)
{
int i;
for(i=1;i<=y;i++)
{
if(!vis[i]&&map[x][i])
{
vis[i]=1;
if(link[i]==-1||dfs(link[i]))
{
link[i]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int c=0;
while(scanf("%d%d%d",&x,&y,&m)!=EOF,x||m||y)
{
int i;
init();
while(m--)
{
int a,b;
scanf("%d%d",&a,&b);
map[a][b]=0;
}
memset(link,-1,sizeof(link));
int ans=0;
for(i=1;i<=x;i++)
{
memset(vis,0,sizeof(vis));
if(dfs(i))
ans++;
}
printf("Case %d: %d\n",++c,x+y-ans);
}
}