Building Block

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1982    Accepted Submission(s): 598

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

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gaojie

题目大意:有n个blocks,定义两种操作，一种是M x y 将block x加入到block y 下面，包括x的下面所有blocks。另一个是C x，计算block  x 下面block 的数量并输出来。

分析：这道题可以通过并查集来解决，首先每个block的跟自己归在一类，，当遇到M操作时，将x 集合加入到y集合下面，并将y的数量加上x的数量，当碰到C 操作时，只要输出相应的数量即可。

1 #include
       
         2 #include
        
          3 #include
         
           4 #define maxlen 30010 5 using namespace std; 6 int father[maxlen],all[maxlen],sum[maxlen]; 7 void Init() 8 { 9     for(int i=0; i