More is better

描述：

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.

输入：

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)

输出：

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

样例输入：

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

样例输出：

4
2

样例解释：

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.

题目大意：

先输入n表示有n组关系，接下去n行每行输入两个正整数a，b表示a和b是朋友，朋友的关系满足传递性，求最大的朋友圈人数。

代码如下：

#include<stdio.h>
#include<string>
int map[10000001],frd[10000001];
int max;
int find(int a)                    //寻找根节点 
{
    if(a!=map[a])                  //如果用return a==map[a]?a:find(map[a])来代替会超时 
    map[a]=find(map[a]);  
    return map[a];
}
void res(int a,int b)                
{
	int q=find(a);
	int w=find(b);
	if(q!=w)						//将w赋值给q的根，此时q的朋友就是w的朋友 
	{
		map[q]=w;
		frd[w]+=frd[q];          
		if(max<frd[w])				 
		max=frd[w];
	}
}
int main()
{
	int t,a,b;
	while(scanf("%d",&t)!=EOF)
	{
		for(int i=0;i<10000001;i++)
		{
			map[i]=i;
			frd[i]=1;
		}
		max=1;									//用max来记录最大朋友数量（包括自己） 
		while(t--)
		{
			scanf("%d %d",&a,&b);
			res(a,b);
		}
		printf("%d\n",max);
	}
	return 0;
}