HDU 1068 Girls and Boys 求独立集（入门）-优快云博客

本文链接：https://blog.youkuaiyun.com/crescent__moon/article/details/9822605

本文探讨了如何通过编程解决大学学生浪漫关系中的最大独立集问题，即找出不受浪漫关系影响的学生的最大数量。通过输入描述学生间浪漫关系的数据，程序计算并输出无法配对的学生人数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5793 Accepted Submission(s): 2592

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

Recommend

JGShining

题意是说男生和女生两两配对。让你求有多少人无法配对。就是求最大独立集，而最大独立集=定点数---最大匹配。因为此题的图是无向图，所以最大匹配还要除以2.如图2，就是此题的图。

#include<stdio.h>
#include<string.h>
using namespace std;
int n;
int link[1007],g[1007][1007];
bool vis[1007];
struct node
{
    int x,y,z;
}box[1007];
bool judge(int i, int j)
{
    if(box[i].x<box[j].x&&box[i].y<box[j].y&&box[i].z<box[j].z)
    return true;
    return false;
}
bool find(int i)
{
    for(int j=0;j<n;j++)
    if(g[i][j]&&!vis[j])
    {
        vis[j]=true;
        if(link[j]==0||find(link[j]))
        {
            link[j]=i;
            return true;
        }
    }
    return false;
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        memset(g,0,sizeof(g));
        memset(link,0,sizeof(link));
        int a,b,num;
        for(int i=0;i<n;i++)
        {
            scanf("%d: (%d) ",&a,&num);
            while(num--)
            {
                scanf("%d",&b);
                g[a][b]=1;
            }
        }
        int count=0;
        for(int i=0;i<n;i++)
        {
            memset(vis,false,sizeof(vis));
            if(find(i))
            count++;
        }
        printf("%d\n",n-count/2);
    }
    return 0;
}