zoj1137 poj1466

Girls and Boys

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Time Limit: 10 Seconds      Memory Limit: 32768 KB

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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 

 最大独立点集：

#include<cstring>
#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
const int maxn = 505;
int match[maxn];int used[maxn];int map[maxn][maxn];int n;
int dfs(int s)
{
    int i,temp;
    for(i=0;i<n;i++)
    {
        if(map[s][i] && !used[i])
        {
            used[i] = 1;
        //    temp = match[i];
            if(match[i]==-1 || dfs(match[i]))
            {
                match[i] = s;
                return 1;
            }
        }
    }
    return 0;
}
int solve()
{
    int i,ans= 0;
    memset(match,-1,sizeof(match));
    for(i=0;i<n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i)) ans++;
        //if(ans==n) break;
    }
    return ans;
}
int main()
{
    int m,x,y,i;
    while(scanf("%d",&n)!=EOF)
    {
        memset(map,0,sizeof(map));
        for(i=0;i<n;i++)
        {
            scanf("%d: (%d)",&x,&m);
            while(m--)
            {
                scanf("%d",&y);
                map[x][y] =  1;
            }
        }
        int t  = solve();
            printf("%d\n",n-t/2);
    }
    return 0;
}