【图算法之二分图】HDU1068---Girls and Boys

最新推荐文章于 2024-04-22 14:46:51 发布

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本文介绍了一种解决最大独立集问题的算法实现，通过输入学生之间的浪漫关系来找出最大的互不相识的学生群体数量。该算法利用了匈牙利算法的思想进行求解。

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Girls and Boys

Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3260 Accepted Submission(s): 1405

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

题目大意：给你每个人互相认识的人，然后问最多能找到多少个人都互不认识。其实就是找：最大独立集合！

#include<stdio.h>
#include<string.h>
int map[501][501],flag[501],pre[501],num;
int get(int i)
{
    int j,k,l;
    for(j=0;j<num;j++)
    {
        if(map[i][j]&&flag[j]==0)
        {
            flag[j]=1;
            if(pre[j]==-1||get(pre[j]))
            {
                pre[j]=i;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    int l,r,i,ll,j,k;
    while(~scanf("%d",&num))
    {
        memset(map,0,sizeof(map));
        memset(pre,-1,sizeof(pre));
        for(i=0;i<num;i++)
        {
            scanf("%d: (%d)", &k, &ll);
            for(j=0;j<ll;j++)
            {
                scanf("%d",&r);
                map[i][r]=1;
            }
        }
        int sum=0;
        for(int i=0;i<num;i++)
        {
            memset(flag,0,sizeof(flag));
            sum+=get(i);
        }
        printf("%d\n",num-sum/2);

    }
    return 0;
}