【HDU1856】More is better（并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

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本文介绍了一种解决最大交友群组问题的算法实现。通过并查集数据结构，该算法可以找出参与项目的最大可能人数。具体操作包括初始化、查找根节点、合并节点等步骤，并最终输出最大群组规模。

这里是题目(～￣▽￣)～

More is better
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 29023 Accepted Submission(s): 10318

Problem Description
Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang’s selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input
The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output
The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input
4
1 2
3 4
5 6
1 6
4
1 2
3 4
5 6
7 8

Sample Output
4
2

Hint
A and B are friends(direct or indirect), B and C are friends(direct or indirect),
then A and C are also friends(indirect).

In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

题：选择互相认识的男孩，最多能选择多少人。需要设出一个数组储存同一组的人数个数最后输出最大的一组。

#include<stdio.h>
const int MAX=1e7+10;
int c[MAX],d[MAX];
int max;
int find(int x)
{
    int p=x;
    while(x!=c[x])
     x=c[x];
    while(c[p]!=x)
    {
        int j=c[p];
        c[p]=x;
        p=j;
    }
    return c[x];
}

void merge(int xx,int yy)
{
    int fx=find(xx);
    int fy=find(yy);
    if(fx!=fy)
    {
        c[fx]=fy;
        d[fy]+=d[fx];
        if(d[fy]>max)
        max=d[fy];
    }
}

int main()
{
    int m,n,a,b;
    while(~scanf("%d",&n))
    {
        max=1;
        for(int i=0;i<=MAX;i++)
        c[i]=i,d[i]=1;            //需要加一个数组计算同组数量
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            merge(a,b);
         } 
         printf("%d\n",max);
    }
return 0;
}