hdoj 1856 more is better （如何记录节点数）

最新推荐文章于 2018-06-14 14:46:08 发布

原创最新推荐文章于 2018-06-14 14:46:08 发布 · 426 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #并查集

并查集专栏收录该内容

3 篇文章

订阅专栏

本文介绍了一道编号为1856的HDU在线评测题目“Moreisbetter”的解决思路。该题涉及1000万个候选人，通过一系列直接朋友关系找出能够形成间接朋友关系的最大群体规模。文章提供了完整的C++代码实现，采用并查集算法进行高效处理。

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1856

More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 19180 Accepted Submission(s): 7054

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

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.

<span style="font-size:14px;">#include<stdio.h>
#include<algorithm>
using namespace std;
int a,b;
int per[10000010];
int n,num[10000010];
int inf;
void init()
{
	for(int i=1;i<=1000000;i++)   //开数组的时候要注意尾标不能大于等于数组长度 
	{
		per[i]=i;
		num[i]=1;  //初始化每个根节点的数量都是1 
	}
}

/*int find(int x){ //路径压缩迭代版本  
    int r = x;  
    while(r!=per[r])   
        r = per[r];//肇东根节点  
    while(x!=r){ //将这棵树上的节点都指向根节点  
        int tmp = per[x];  
       per[x] = r;  
        x= tmp;  
    }  
    return r;  
}*/ 

int find(int x)
{
	if(x==per[x])
	return x;
	else
	return per[x]=find(per[x]);
}
int join(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		per[fx]=fy;
		num[fy]+=num[fx];  //记录根节点的上节点数量，总把子节点的数量加在父节点上		
	}
	if(num[fy]>inf) 
	inf=num[fy];   //不断保存父节点的节点数最大值 
}
int main()
{
	int i;
	while(scanf("%d",&n)!=EOF)
	{
		init();
		inf=1;
		for(i=1;i<=n;i++)		
	   {
	   		scanf("%d%d",&a,&b);
	   		join(a,b);
	   }
	   
		printf("%d\n",inf);
		
	}
}</span>