HDU 1856 简单并查集

并查集解决朋友圈问题

最新推荐文章于 2021-03-19 13:26:19 发布

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本文介绍了一道关于使用并查集算法解决朋友圈问题的题目。该问题涉及到10000000个人中找出最大的朋友圈规模。通过实现特定的并查集算法，包括路径压缩优化，最终找到最优解。

第一次做并查集，MARK。。。这是一道简单的并查集题目，（连我都能做出来，当然简单了），题目大意是，有10000000人排队来，他们当中两个两个是朋友，朋友的朋友也是朋友，（把他们归为一个集合）；输入是：n，接下来n组数据，每组a，b代表a和b是朋友，输出最大朋友圈的人数；

题目：More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 7376 Accepted Submission(s): 2708

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

#include<iostream>
#include<cstdio>
#define MAXNUM 10000010
using namespace std;
int set[MAXNUM],num[MAXNUM];
int find(int x)
{
    if(set[x] == x)return x;
    else
     return set[x] = find(set[x]);
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        for(int i = 1; i < MAXNUM; i++)
        {
            set[i] = i;
            num[i] = 1;
        }
        for(int i = 1; i <= n; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            int fx = find(x);
            int fy = find(y);
            if(fx != fy)
            {
                set[fx] = fy;
                num[fy] += num[fx];
            }
        }
        int max = 1;
        for(int i = 1; i < MAXNUM; i++)
        {
            if(set[i]==i && max < num[i])
            max = num[i];
        }
        printf("%d",max);
    }
    return 0;
}

精华在于find()函数；

int find(int x)
{
    if(set[x] == x)return x;
    else
     return set[x] = find(set[x]);
}

find函数作用是压缩路径；这样能减少查找的时候的步骤；