1466 Girls and Boys//MaxMatch

Girls and Boys
Time Limit: 5000MS
 
Memory Limit: 10000K
Total Submissions: 5324
 
Accepted: 2304
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

Source
Southeastern Europe 2000
2 를 나누는 것을 잊지 마라
남녀 학생의 수가 정해져 있지 않기 때문에 중복 계산이 있다
그래서 2를 제외하고는 중복 계산을 여과했다.
 
 
#include
#include
bool mat[505][505];
bool usedif[505];
int link[505];
int n;
bool can(int t)
{
     int i;
     for(i=0;i      if(!usedif[i]&&mat[t][i])
      {
         usedif[i]=true;
         if(link[i]==-1||can(link[i]))
         {
            link[i]=t;
            return true;
         }
      }
     return false;
}
 
int maxMatch()
{
    int i,num=0;
    memset(link,-1,sizeof(link));
    for(i=0;i    {
        memset(usedif,0,sizeof(usedif));
        if(can(i))
          num++;
    }
    return num;
}
int main()
{
    int i,u,v,m,ans;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
           break;
        memset(mat,0,sizeof(mat));
        for(i=0;i        {
           scanf("%d: (%d)",&u,&m);//주의//여기 입력 주의
           while(m--)
           {
               scanf("%d",&v);
               mat[u][v]=true;
           }
        }
        ans=n-maxMatch()/2;
        printf("%d/n",ans);
    }
    return 0;
}