hdu 1856 More is better 并查集

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.

简单并查集，只是要求最大树中元素的个数；

注意：n=0时输出1；还有tle~..~

#include<cstdio>#include<cmath>#include<iostream>#include<cstring>using namespace std;int a[10000005],sum[10000005];int f(int x){    return a[x]==x?x:a[x]=f(a[x]);//这个貌似可以减少一半的查找时间}int max(int x,int y){    return x>y?x:y;}int main(){    int n;    while(~scanf("%d",&n))    {        if(n==0)        {            printf("1\n");            continue;        }//注意判断；        int i,x,y;        for(i=1; i<=10000000; i++)        {            a[i]=i;            sum[i]=1;        }        int su=0,maxn=0;        for(i=0; i<n; i++)        {            scanf("%d%d",&x,&y);            int xx=f(x);            int yy=f(y);            if(xx!=yy)                sum[xx]=sum[xx]+sum[yy];            a[yy]=xx;            maxn=max(maxn,max(x,y));        }        for(i=1; i<=maxn; i++)        {            if(sum[i]>su)                su=sum[i];        }        printf("%d\n",su);    }}