gin main 並查集 keep better sele scan ted dia

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

大意：

朋友圈問題，A和B是朋友，B和C是朋友則A和C也是朋友，依次類推，題目的意思就是求最大的朋友圈，即求最大集合中元素的個數。裸的並查集加個秩數組就行了。

註意當朋友對為0時要特判一下，因為題目中寫的很清楚0<=n<=1000000。

#include<iostream>  
using namespace std;  
const int MAX=10000005;
int pre[MAX],rank[MAX],maxx;
int find(int r)                               //
 
找最上級朋友
{  
    if(pre[r]!=r)
    pre[r]=find(pre[r]);
    return pre[r];
}   
void join(int x,int y)                       
{  
    int fx=find(x),fy=find(y);                //加入朋友圈
    if(fx!=fy)  
    {
        pre[fx]=fy;  
        rank[fy]+=rank[fx];                   //朋友圈數量增加
        if(rank[fy]>maxx)  
            maxx=rank[fy];                    //判斷朋友數量的最大值
    }
}
int main()  
{  
    int i,n,x,y;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        {
            printf("1\n");
            continue;    
        }
        for(i=0;i<MAX;++i)                    //rank、pre數組初始化
        {  
            pre[i]=i;   
            rank[i]=1;             
        } 
        maxx=-1;                              //最大值初始化
        for(i=0;i<n;++i)   
        {  
            scanf("%d%d",&x,&y);
            join(x,y);
        }  
        printf("%d\n",maxx);    
    }
    return 0;  
} 

【HDU 1856】More is better （裸並查集+記錄秩）