hdoj 1856 More is better【求树的节点数】

More is better

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 17683    Accepted Submission(s):
6493

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.

 

 并查集求相连的节点的节点数

#include<stdio.h>
#include<string.h>
#define MAX 10000010
#define maxn(a,b)(a>b?a:b)
int set[MAX],path[MAX];
int a[100010],b[100010];
int sum=0;
int find(int fa)
{
	int t;
	int ch=fa;
	while(fa!=set[fa])
	fa=set[fa];
	while(ch!=fa)
	{
		t=set[ch];
		set[ch]=fa;
		ch=t;
	}
	return fa;
}
void mix(int x,int y)
{
	int fx,fy;
	fx=find(x);
	fy=find(y);
	if(fx!=fy)
	{
		set[fx]=fy;
		path[fy]+=path[fx];
		if(sum<path[fy])
		sum=path[fy];
	}
}
int main()
{
	int n,m,j,i,max,point;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
		{
			printf("1\n");
			continue;
		}
		max=0;
	    for(i=1;i<=n;i++)
	    {
	    	scanf("%d%d",&a[i],&b[i]);
	    	if(max<maxn(a[i],b[i]))
	    	max=maxn(a[i],b[i]);
	    }
	    for(i=1;i<=max;i++)
	    {
	    	set[i]=i;
	    	path[i]=1;
	    }
	    for(i=1;i<=n;i++)
	    {
	    	mix(a[i],b[i]);
	    }
	    printf("%d\n",sum);
	    sum=0;
	}
	return 0;
}