标签：poj
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 4920 | Accepted: 2399 |
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
题意：幼儿园里男孩纸跟男孩纸认识，女孩纸跟女孩纸认识，男孩纸跟女孩纸也部分认识。
需要求出最大的，能使里面所有男孩纸女孩纸都认识的集合（最大团）
1.独立集：任意两点都不相连的顶点的集合
2.定理：最大独立集=顶点数-最大匹配边
3.完全子图：任意两点都相连的顶点的集合（最大完全子图即最大团）
4.定理：最大团=原图补图的最大独立集=顶点数-最大匹配数
注意，要反向建图（把认识的变成不认识的，不认识的成为认识的，这样才能求补图）
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int M = 1000 + 5;
int k, m, n;
int link[M];
bool MAP[M][M];
bool cover[M];
int ans;
int cas=0;
void init()
{
int x, y;
memset(MAP, true, sizeof(MAP));
for(int i=1; i<=k; i++)
{
scanf("%d%d", &x, &y);
MAP[x][y]=false;
}
}
bool dfs(int x)
{
for(int y=1; y<=m; y++)
{
if(MAP[x][y] && !cover[y])
{
cover[y]=true;
if(!link[y] || dfs(link[y]))
{
link[y]=x;
return true;
}
}
}
return false;
}
int main()
{
while(scanf("%d%d%d", &n, &m, &k) && n && m && k)
{
cas++;
ans=0;
init();
memset(link, false, sizeof(link));
for(int i=1; i<=n; i++)
{
memset(cover, 0, sizeof(cover));
if( dfs(i) )
ans++;
}
printf("Case %d: %d\n", cas,(n+m)-ans);
}
return 0;
}
POJ 3692：Kindergarten（最大团）,布布扣,bubuko.com
标签：poj
原文地址：http://blog.csdn.net/u013487051/article/details/37662739
