九度 题目1444：More is better

时间限制：3 秒

内存限制：100 兆

特殊判题：否

提交：913

解决：423

题目描述：

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：

41 23 45 61 641 23 45 67 8

样例输出：

42

用并查集求得每个集合的元素的个数，最后输出最大的集合的元素个数。合并两个集合时，也原集合的元素个数累加到新集合中。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int r[10000003],c[10000003];
int Findroot(int x)
{
    if(r[x]==-1) return x;
    else return Findroot(r[x]);
}
int main()
{
   int n;
   while(scanf("%d",&n)!=EOF)
   {
       int sum=0;
       for(int i=1;i<=10000001;i++)
       {
           r[i]=-1;
           c[i]=1;
       }
      for(int i=1;i<=n;i++)
      {
          int a,b;
          scanf("%d%d",&a,&b);
          a=Findroot(a);
          b=Findroot(b);
          if(a!=b)
          {
              r[a]=b;
              c[b]+=c[a];
          }
      }
      for(int i=1;i<=10000000;i++)
          if(r[i]==-1&&c[i]>sum)
          sum=c[i];
      printf("%d\n",sum);
   }
   return 0;
}