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
题意概括：
一群人做过山车，一排2个座位，男生女生配一对，但女生只想跟认识的人配一对，问有多少人配不成对。
解题思路：
二分匹配问题，最大独立集=总人数-最大匹配数/2;
代码：

#include<stdio.h>
#include<string.h>
#define maxn 1010
int map[maxn][maxn],bk[maxn],ma[maxn];
int k;
int dfs(int d)
{
    for(int i=0;i<k;i++)//为k找关系 
    {
        if(bk[i]==0&&map[d][i])//如果这两个人认识，且i在这次匹配中没被找过 
        {
            bk[i]=1;//标记为已找过 
            if(ma[i]==0||dfs(ma[i]))//如果i没有人有关系，或者i的关系人能找到新的关系就把i,d两人建立关系 
            {
                ma[i]=d;
                return 1;
            }

        }
    }
    return 0;
}
int main()
{
    int t;
    while(scanf("%d",&t)!=EOF)
    {
        memset(map,0,sizeof(map));//初始化男女关系 
        int x,y,n;
        k=t;
        while(t--)//建立男女关系 
        {
            scanf("%d: (%d)",&x,&n);
            while(n--)
            {
                scanf("%d",&y);
                map[x][y]=1;
            }
        }
        int s=0;
        memset(ma,0,sizeof(ma));//假设都没有找到关系 
        for(int i=0;i<k;i++)
        {
            memset(bk,0,sizeof(bk));//为此人找关系前假设所有人都没有关系 
            if(dfs(i))//假如建立关系 
            s++;
        }
        printf("%d\n",k-s/2);
    }
    return 0;
}