Building Block

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 151   Accepted Submission(s) : 57

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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

题意：

M x y把x的管放在y管的上面

C x求x的下面有多少个砖块

down[i]表示i到根节点之间有多少块砖，再维护一个sum[x]数组，表示管x上一共串有多少个砖块

find每次更新down数组；

#include<iostream>
#include<stdio.h>
using namespace std;
int par[30005];
int sum[30005];//上面有多少；
int down[30005];
int findi(int x)
{
    if(x==par[x])
        return x;
    int p=findi(par[x]);
    down[x]=down[par[x]]+down[x];
    return par[x]=p;
}
void unioni(int x,int y)
{
    int xx=findi(x);
    int yy=findi(y);
    if(xx!=yy)
    {
        par[xx]=yy;
        down[xx]=sum[yy];
        sum[yy]=sum[yy]+sum[xx];

    }

}

int main()
{
int n;
for(int i=0;i<30005;i++)
    sum[i]=1;
for(int i=0;i<30005;i++)
    down[i]=0;
for(int i=0;i<30005;i++)
     par[i]=i;
scanf("%d",&n);

while(n--)
{
    char a;
    cin>>a;
    if(a==‘M‘)
    {
        int b,c;
       scanf("%d%d",&b,&c);
        unioni(b,c);
    }
    if(a==‘C‘)
    {
        int d;
         scanf("%d",&d);
         findi(d);
         printf("%d\n",down[d]);
    }
}

    return 0;
}