POJ 1466 Girls and Boys (最大独立点集)

最新推荐文章于 2019-03-13 00:01:02 发布

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本文探讨了一种求解最大独立点集问题的算法，并通过一个具体的编程实例展示了如何运用最大匹配算法来解决该问题。具体地，文章介绍了一个大学研究项目背景下的问题：寻找最大的学生集合，使得此集合中任意两人未曾有过浪漫关系。通过对输入数据进行处理并使用图论中的最大匹配算法，最终计算出满足条件的学生数量。

Girls and Boys

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 9151		Accepted: 4046

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

/*
最大独立点集+最大匹配=n+m,
这里m=n，所以ans=(2*n-maxf)/2=n-maxf/2;
*/
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int maxn=501;
vector<int>edge[maxn];
bool vis[maxn];
int match[maxn],n;
bool find(int now)
{
    int i,v;
    for(i=0; i<edge[now].size(); ++i)
        if(!vis[v=edge[now][i]])
        {
            vis[v]=true;
            if(match[v]==-1||find(match[v]))
            {
                match[v]=now;
                return true;
            }
        }
    return false;
}
int main()
{
    int stu_id,love_stu,love_num,i,j;
    while(scanf("%d",&n)!=-1)
    {
        for(i=0; i<n; i++)
        {
            edge[i].clear();
            match[i]=-1;
        }
        for(i=0; i<n; i++)
        {
            scanf("%d: (%d)",&stu_id,&love_num);
            while(love_num--)
            {
                scanf("%d",&love_stu);
                edge[stu_id].push_back(love_stu);
            }
        }
        int maxf=0;
        for(i=0; i<n; i++)
        {
            for(j=0; j<n; j++) vis[j]=false;
            if(find(i)) maxf++;
        }
        printf("%d\n",n-maxf/2);
    }
    return 0;
}