HDU——T 2818 Building Block

http://acm.hdu.edu.cn/showproblem.php?pid=2818

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5935    Accepted Submission(s): 1838

Problem Description

John are playing with blocks. There are N blocks (1 <= N <= 30000) numbered 1...N。Initially, there are N piles, and each pile contains one block. Then John do some operations P times (1 <= P <= 1000000). There are two kinds of operation:

M X Y : Put the whole pile containing block X up to the pile containing Y. If X and Y are in the same pile, just ignore this command. 
C X : Count the number of blocks under block X 

You are request to find out the output for each C operation.

 

 

Input

The first line contains integer P. Then P lines follow, each of which contain an operation describe above.

 

 

Output

Output the count for each C operations in one line.

 

 

Sample Input

6 M 1 6 C 1 M 2 4 M 2 6 C 3 C 4

 

 

Sample Output

1 0 2

 

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

 

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多组数据、、貌似因为这个WA好多次了、、

#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

const int N(30005);
int fa[N],sum[N],beh[N];

int find(int x)
{
    if(fa[x]==x) return x;
    int dad=fa[x];
    fa[x]=find(fa[x]);
    beh[x]+=beh[dad];
    return fa[x];
}
void combine(int x,int y)
{
    x=find(x),y=find(y);
    if(x==y) return ;
    beh[x]=sum[y];
    sum[y]+=sum[x];
    fa[x]=y;
}
void init()
{
    for(int i=0;i<N;i++) fa[i]=i,sum[i]=1,beh[i]=0;
}

int main()
{
    for(int p,u,v;~scanf("%d",&p);)
    {
        init();
        for(char ch[2];p--;)
        {
            scanf("%s%d",ch,&u);
            if(ch[0]=='M')
            {
                scanf("%d",&v);
                combine(u,v);
            }
            else find(u),printf("%d\n",beh[u]);
        }
    }
    return 0;
}