Girls and Boys

最新推荐文章于 2022-04-19 22:03:08 发布

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最新推荐文章于 2022-04-19 22:03:08 发布

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分类专栏：二分图

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本文介绍了一个基于最大匹配算法的问题——如何从一组可能存在浪漫关系的学生中找出最大数量的互不相关的个体集合。通过将问题转化为图论中的二分图匹配问题，并使用匈牙利算法进行求解，提供了一段具体的C++实现代码。

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Description

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.

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, for n subjects.
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

分析：

题意是让我们找出一个最大的集合使得该集合的任意两个人没有关系，所以就等于人数减去匹配的数目；

注意的是该题没有给出哪些为男哪些为女，所以就把一人拆成两人来算，最终匹配的数目要/2；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[50005];
bool fr[505][505];
bool vis[505];
int k,n,m;
bool find(int x)
{
	for(int i=0;i<n;i++)
	{
		if(fr[x][i]&&!vis[i])
		{
			vis[i]=true;
			if(f[i]==-1)
			{
				f[i]=x;
				return true;
			}
			else if(find(f[i]))
			{
				f[i]=x;
				return true;
			}
		}
	}
	return false;
}
int main()
{         int t,a,b;
		while(scanf ("%d",&t)!=EOF)
		{ 
	   memset(f,-1,sizeof(f));
	   memset(fr,false,sizeof(fr));
		int y;
		 n=t;
	     while(t--)
	     {   scanf("%d: (%d) ",&a,&b);
	         for(int i=0;i<b;i++)
	         {
			 scanf("%d",&y);
			 fr[a][y]=true;
		     }
	     }
		int ans=0;
		for(int i=0;i<n;i++)
		{     
           memset(vis,false,sizeof(vis));
           if(find(i))
           ans++;
		}
		printf("%d\n",n-ans/2);
}
	return 0;
}