POJ 3692：Kindergarten 求补图的最大点独立集 头一次接触这样的做法

Kindergarten

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5884		Accepted: 2877

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

题意是幼儿园有很多小孩，男生和男生互相熟悉，女生和女生互相熟悉。然后给出了一些男生和一些女生互相熟悉的关系，要找到一个最大的群体，这个群体中所有人都互相熟悉，问这个群体最多多少人。

二分图的最大点独立集(二分图中两两互不相连的点的集合) = N - 二分图的最大匹配数量

这个题要求的点独立集是要两两互相连，所以这要求的是补图的最大点独立集。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

int grid[205][205];
int link[205];
int visit[205];
int n,m,k,V1,V2;
int result;

bool dfs(int x)
{
	int i;
	for(i=1;i<=V2;i++)
	{
		if(grid[x][i]==0&&visit[i]==0)
		{
			visit[i]=1;
			if(link[i]==-1||dfs(link[i]))
			{
				link[i]=x;
				return true;
			}
		}
	}
	return false;
}

void Magyarors()
{
	int i;

	result=0;
	memset(link,-1,sizeof(link));//！！这里不能是0

	for(i=1;i<=V1;i++)
	{
		memset(visit,0,sizeof(visit));
		if(dfs(i))
			result++;
	}
	cout<<V1+V2-result<<endl;
}

int main()
{
	int i,j,pair,temp1,temp2;

	for(i=1;;i++)
	{
		scanf("%d%d%d",&n,&m,&pair);
		V1=n;
		V2=m;
		if(n==0&&m==0&&pair==0)
			break;

		cout<<"Case "<<i<<": ";
		memset(grid,0,sizeof(grid));

		for(j=1;j<=pair;j++)
		{
			scanf("%d%d",&temp1,&temp2);
			grid[temp1][temp2]=1;
		}
		Magyarors();
	}
	return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。