春季补题二分图I

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1137
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.

An example is given in Figure 1.

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

Output

5
2
告诉你学生总人数和每个学生与哪些人有关系找出最大数量学生使其中没有两个学生有关系
答案为总人数减去最小覆盖点数，而最小覆盖点数等于最大匹配数，故用二分图求解

#include<stdio.h>
#include<string.h>
#define N 505
using namespace std;
int n,vis[N],map[N][N],pre[N];
int dfs(int x)
{
	for(int i=0;i<n;i++)
	{
		if(map[x][i]&&!vis[i])
		{
			vis[i]=1;
			if(pre[i]==0||dfs(pre[i]))
			{
				pre[i]=x;
				return 1;
			}			
		}
	}
	return 0;
}
int main()
{
	while(~scanf("%d",&n))
	{
		memset(map,0,sizeof(map));
		memset(pre,0,sizeof(pre));
		for(int i=0;i<n;i++)
		{
			int num,k;
            scanf("%d: (%d)",&k,&num);
            for(int j=0;j<num;j++)
			{
				int temp;
                scanf("%d",&temp);
                map[i][temp]=1;
            }
        }
        int ans=0;
		for(int i=0;i<n;i++)
	    {
		    memset(vis,0,sizeof(vis));
		    if(dfs(i)==1)
		    ans++;
	    }
	    printf("%d\n",n-ans/2);
	}
}