More is better

Time Limit:1000MS     Memory Limit:102400KB     64bit IO Format:%I64d & %I64

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. 

题意：朋友和朋友住在一起，不是朋友的不能住在一起，A与B是朋友,B与C是朋友，则A与C也是朋友，问最大朋友数。
分析：典型并查集。

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;

int s[10000005];
int sum[10000005];
int maxn;

int find(int x)
{
    return s[x]==x?x:find(s[x]);
}

void init ()
{
    maxn=1;//注意这里最小人数一定是 1
    for(int i=0;i<=10000000;i++)
    {
        s[i]=i;
        sum[i]=1;
    }
}

void merge(int a,int b)
{
    a=find(a);
    b=find(b);
    if(a==b) return;
    if(sum[a]>=sum[b])
    {
        s[b]=a;
        sum[a]+=sum[b];
        if(maxn<sum[a]) maxn=sum[a];
    }
    else
    {
        s[a]=b;
        sum[b]+=sum[a];
        if(maxn<sum[b]) maxn=sum[b];
    }
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        init();//初始化并查集
        while(n--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            merge(a,b);
        }
        printf("%d\n",maxn);
    }
}