HDU1068/POJ1466_Girls and Boys(二分图/最大独立集=N-最大匹配)

解题报告

http://blog.csdn.net/juncoder/article/details/38160591

题目传送门（POJ）

题目传送门（HDU）

题意：

求满足条件的最大集合：集合内不论什么两个人都没有浪漫关系

思路：

跟POJ2771一样的题，变的简单多了。POJ2771解题报告

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,mmap[550][550],pre[550],vis[550];
int dfs(int x)
{
    for(int i=0; i<n; i++) {
        if(!vis[i]&&mmap[x][i]) {
            vis[i]=1;
            if(pre[i]==-1||dfs(pre[i])) {
                pre[i]=x;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int i,j,a,b,k;
    while(~scanf("%d",&n)) {
        memset(mmap,0,sizeof(mmap));
        memset(pre,-1,sizeof(pre));
        for(i=1; i<=n; i++) {
            scanf("%d: (%d) ",&a,&k);
            for(j=1; j<=k; j++) {
                scanf("%d",&b);
                mmap[a][b]=1;
            }
        }
        int ans=0;
        for(i=0; i<n; i++) {
            memset(vis,0,sizeof(vis));
            ans+=dfs(i);
        }
        printf("%d\n",n-ans/2);
    }
    return 0;
}

Girls and Boys

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