hdu2458 2010.3.6

最新推荐文章于 2018-08-10 09:11:05 发布

Batman_

最新推荐文章于 2018-08-10 09:11:05 发布

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本文探讨了一个幼儿园游戏选拔问题，即如何从已知相互认识关系的孩子中挑选最多数量的孩子参与游戏，确保所有参与者彼此认识。该问题通过二分图匹配算法解决，采用一种特殊的建图方式：将没有直接关系的男孩和女孩之间建立连接。

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hdu2458 2010.3.6

二分图，建图的方式很特别，会想不到——没有关系的建一条边

Kindergarten

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 111 Accepted Submission(s): 58

Problem 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. Inaddition to that, some girls and boys know each other. Now the teachers want topick some kids to play a game, which need that all players know each other. Youare 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 whoknow each other, respectively.

Each of the following M lines contains twointegers 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 theboys are numbered from 1 to B.

The last test case is followed by a linecontaining three zeros.

Output

For each test case, print a line containingthe test case number( beginning with 1) followed by a integer which is themaximum 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

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#include <stdio.h>
#include <string.h>

#define MAX 210

char map[MAX][MAX],yt[MAX];
int g,b,y[MAX];

int judge(int);

int main(void)
{
	int m,i,p,q,t,sum;

	t=0;
	scanf("%d %d %d",&g,&b,&m);
	while (g!=0 || b!=0 || m!=0)
	{
		t++;
		memset(map,0,sizeof(map));

		while (m>0)
		{
			scanf("%d %d",&p,&q);
			map[p][q]=1;
			m--;
		}

		sum=0;
		memset(y,0,sizeof(y));
		for (i=1;i<=g;i++)
		{
			memset(yt,0,sizeof(yt));
			if (judge(i))  sum++;
		}

		printf("Case %d: %d\n",t,g+b-sum);

		scanf("%d %d %d",&g,&b,&m);
	}

	return 0;
}

int judge(int a)
{
	int i;

	for (i=1;i<=b;i++)
	{
		if (!map[a][i] && !yt[i])
		{
			yt[i]=1;
			if (y[i]==0 || judge(y[i]))
			{
				y[i]=a;
				return 1;
			}
		}
	}
	return 0;
}