poj 1466 Girls and Boys

最新推荐文章于 2018-10-16 16:16:17 发布

风吹过的烟花

最新推荐文章于 2018-10-16 16:16:17 发布

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分类专栏： ACM算法_二分图文章标签： poj 二分图

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本文介绍了一个基于匈牙利算法求解最大独立集的问题，通过两两匹配的方式找到学生间无浪漫关系的最大群体数量。该算法首先读取输入的学生数量及他们之间的关系，然后使用匈牙利算法计算最大匹配数，并据此得出最大独立集的大小。

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Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 13135		Accepted: 5819

Description

In the second year of the university somebody started a study on the romantic relations between the students. The relation "romantically involved" is defined between one girl and one boy. For the study reasons it is necessary to find out the maximum set satisfying the condition: there are no two students in the set who have been "romantically involved". The result of the program is the number of students in such a set.

Input

The input contains several data sets in text format. Each data set represents one set of subjects of the study, with the following description:

the number of students
the description of each student, in the following format
student_identifier:(number_of_romantic_relations) student_identifier1 student_identifier2 student_identifier3 ...
or
student_identifier:(0)

The student_identifier is an integer number between 0 and n-1 (n <=500 ), for n subjects.

Output

For each given data set, the program should write to standard output a line containing the result.

Sample Input

7
0: (3) 4 5 6
1: (2) 4 6
2: (0)
3: (0)
4: (2) 0 1
5: (1) 0
6: (2) 0 1
3
0: (2) 1 2
1: (1) 0
2: (1) 0

Sample Output

5
2

题目大意：有n个学生，每一个人，都有他认识的人，找出最大的人数，使这些人没关系。

最大独立集=点数-最大匹配数

两两匹配，除2

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int N=505;
int cost[N][N];
int link[N],used[N];
int n;
int dfs(int x)
{
	for(int i=0;i<n;i++)
	{
		if(!used[i]&&cost[x][i])
		{
			used[i]=1;
			if(link[i]==-1||dfs(link[i]))
			{
				link[i]=x;
				return true;
			}
		}
	}
	return false;
}
int hungary()
{
	int num=0;
	memset(link,-1,sizeof(link));
	for(int i=0;i<n;i++)
	{
		memset(used,0,sizeof(used));
		if(dfs(i))
			num++;
	}
	return n-num/2;
}
int main()
{
	int cnt;
	int a,b;
	while(scanf("%d",&n)!=EOF)
	{
		memset(cost,0,sizeof(cost));
		for(int i=0;i<n;i++)
		{
			scanf("%d: (%d)",&a,&cnt);
			for(int j=0;j<cnt;j++)
			{
				scanf("%d",&b);
				cost[a][b]=1;
			}
		}
		printf("%d\n",hungary());
	}
	return 0;
}