HDOJ题目1068Girls and Boys（二分图最大独立集，匈牙利算法模板）

本文链接：https://blog.youkuaiyun.com/yu_ch_sh/article/details/38580959

本文讨论了如何解决二分图中的最大独立集问题，通过构建二分图并利用深度优先搜索算法来确定最大匹配数，最终计算出最大独立集的数量。详细介绍了输入数据的格式和程序实现细节。

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Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7495 Accepted Submission(s): 3426

Problem 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

Source

Southeastern Europe 2000

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题目理解一下就是求二分图的最大独立集。而二分图的最大独立集数=节点数（n）— 最大匹配数（m）。所以关键在于求二分图的最大匹配数。

分析：题目是明显的二分图模型，但是输入没有指明那些是男孩/女孩。所以要拆点，把 i 拆成 i 跟 i' 分别放入X、Y，构造二分图。原先U = V - M，所以 2U = 2V - 2M。分析：

故U = n - M'/2。

ac代码

#include<stdio.h>
#include<string.h>
int map[1010][1010],v[100000],link[100000];
int n;
int dfs(int u)
{
	int i;
	for(i=0;i<n;i++)
	{
		if(!v[i]&&map[u][i])
		{
			v[i]=1;
			if(!link[i]||dfs(link[i]))
			{
				link[i]=u;
				return 1;
			}
		}
	}
	return 0;
}
int main()
{
	while(scanf("%d",&n)!=EOF)
	{
		int i,sum=0;
		memset(map,0,sizeof(map));
		memset(link,0,sizeof(link));
		for(i=1;i<=n;i++)
		{
			int u,num;
			scanf("%d: (%d) ",&u,&num);
			while(num--)
			{
				int vm;
				scanf("%d",&vm);
				map[u][vm]=1;
			}
		}
		for(i=0;i<n;i++)
		{
			//if(!link[i])//不要加这个
			//{
				memset(v,0,sizeof(v));
				if(dfs(i))
					sum++;
			//}
		}
		printf("%d\n",n-sum/2);
	}
}