Girls and Boys 【最大独立集团】

最新推荐文章于 2021-07-11 14:26:04 发布

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最新推荐文章于 2021-07-11 14:26:04 发布

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分类专栏：最大匹配+独立集+团+最小路径.点覆盖

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本文介绍了一种通过最大匹配算法来解决寻找学生间最大独立集合的问题。该问题来源于大学研究背景下，目标是找出最大的学生集合，使得集合内任意两人之间没有浪漫关系。文章提供了完整的C语言实现代码，并通过示例输入输出进行验证。

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Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11291 Accepted Submission(s): 5293

Problem Description
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.

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
思路：找最大独立集合 ==图中点的个数 – 最大匹配数目
代码

#include<stdio.h>
#include<string.h>
#include<math.h>
#define M 1000+110
#define MOD 100000007
#define LL long long
int used[M];
int mapp[M][M];
int pipei[M];
    int n;
int find(int x)
{
    int i,j;
    for(i=0;i<n;i++)
    {
        if(!used[i]&&mapp[x][i])
        {
                used[i]=1;
            if(pipei[i]==-1||find(pipei[i]))
            {
                pipei[i]=x;
                return 1;
            }
        }
    }
    return 0;
 } 
int main()
{

    while(~scanf("%d",&n))
    {
        int i,j;
        memset(mapp,0,sizeof(mapp));
        memset(pipei,-1,sizeof(pipei));
        for(i=0;i<n;i++)
        {
            int a,num;
            scanf("%d: (%d)",&a,&num);
            while(num--)
            {
                int b;
                scanf("%d",&b);
                mapp[a][b]=1;  // 单向的 
            }
        }
        int sum=0;
        for(i=0;i<n;i++) 
        {
            memset(used,0,sizeof(used));
            if(find(i)) sum++;
         } 

         printf("%d\n",n-sum/2);  //最大独立集团 的个数 
    } // 因为这个是 一个序列 和自身序列的匹配肯定会重复 ，

    return 0;
 }