hdu 1856 More is better

最新推荐文章于 2020-02-23 10:21:17 发布

zz09

最新推荐文章于 2020-02-23 10:21:17 发布

阅读量720

点赞数

CC 4.0 BY-SA版权

分类专栏：并查集文章标签：并查集

本文链接：https://blog.youkuaiyun.com/infinitezhen/article/details/8870563

并查集专栏收录该内容

8 篇文章

订阅专栏

本文探讨了如何在大规模人群中高效筛选出最佳团队成员，通过分析朋友关系网络，实现复杂项目的人力资源配置优化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

More is better

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.

代码：

#include<stdio.h>
#define MAX 10000002

int father[MAX];
int rank[MAX];
void Make_Set(void)
{
int i;
for(i=1;i<=MAX;i++)
{
father[i]=i;
rank[i]=1;
}
}

int Find_Set(int x)//查找x元素所在的集合，回溯时压缩路径
{
if(x!=father[x])
{
father[x]=Find_Set(father[x]);//这个回溯的压缩路径是精华
}
return father[x];
}

void Union(int x,int y)//合并
{
x=Find_Set(x);
y=Find_Set(y);
if(x!=y)
{
father[x]=y;
rank[y]+=rank[x];
}
}

int main()
{
int t,x,y,i,n,m;
while(scanf("%d",&t)!=EOF)
{
Make_Set();
int max=0;
while(t--)
{

scanf("%d%d",&n,&m);
Union(n,m);
}
for(i=1;i<=MAX;i++)
{
if(father[i]==i)
{
if(rank[i]>max)
max=rank[i];
}
}
printf("%d\n",max);
}
return 0;
}