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傳送門 Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 12717		Accepted: 5659

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 【解析】 題意 在一群有戀愛關系的男女中選出兩兩之間沒有戀愛關系的人，使人數最多。 方法：二分圖求最大獨立集，最大獨立集=點數-最大匹配。 沒有戀愛關系說明沒有連邊，就是求最大獨立集。最大獨立集中的點兩兩之間沒有連邊。 由於性別未知，所以求的最大匹配/2。因為原來二分圖是男的一邊女的一邊，而現在二分圖左邊是所有人右邊也是所有人，所以要/2‘ 註意：學生的標號從0開始，所以for(int i=0;i<n;i++). 【code】

#include<iostream>
#include
 
<cstdio>
#include<cstring>
using namespace std;
#define N 500+10
int map[N][N],match[N];
int cnt,n,m,id,num,po;
bool vis[N];
bool path(int x)
{
    for(int i=0;i<n;i++)
    {
        if(!vis[i]&&map[x][i])
        {
            vis[i]=1;
            if(!match[i]||path(match[i]))
            {
                match[i]=x;
                return true;
            }
        }
    }
    return false;
}
int main()
{
    while(cin>>n)
    {
        cnt=0;
        memset(map,0,sizeof(map));
        memset(match,0,sizeof(match));
        for(int i=1;i<=n;i++)
        {
            scanf("%d: (%d)",&id,&num);
            for(int j=1;j<=num;j++)
            {
                scanf("%d",&po);
                map[id][po]=1;
            }
        }
        for(int i=0;i<n;i++)
        {
          memset(vis,0,sizeof(vis));
          if(path(i))cnt++;
        }
        cout<<n-cnt/2<<endl;
    }
    return 0;
} 

Girls and Boys