hdu1856(基础并查集)

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

41 23 45 61 641 23 45 67 8

 

Sample Output

42
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：求联通块里面元素个数最多。
思路：并查集将这些人分再不同集合里面，最后直接遍历寻找。
代码：
#include<bits/stdc++.h>using namespace std;const int maxn=1e7+7;int father[maxn],vis[maxn];int findset(int x){    if(x!=father[x])        father[x]=findset(father[x]);    return father[x];}void Union(int a,int b){    int fa=findset(a);    int fb=findset(b);    if(fa==fb) return;    father[fa]=fb;    vis[fb]+=vis[fa];}int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i=1;i<maxn;i++)        {            father[i]=i;            vis[i]=1;        }        for(int i=1;i<=n;i++)        {            int x,y;            scanf("%d%d",&x,&y);            Union(x,y);        }        int ans=0;        for(int i=1;i<maxn;i++)        {            if(father[i]==i)            {                if(vis[i]>ans)                {                    ans=vis[i];                }             }        }        printf("%d\n",ans);    }    return 0;}