POJ 1988 Cube Stacking

Description

Farmer John and Betsy is playing a game with N (1 <= n <= 30,000) identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= p <= 100,000) operation. There is types of operations:
Moves and counts.
* In a move operation, Farmer John asks Bessie to move the stack containing cube X on top of the stack containing cube Y.
* In a count operation, Farmer John asks Bessie to count the number of cubes on the stack with Cube X is under the C Ube X and report that value.

Write a program that can verify the results of the game.

Input

* Line 1: A single integer, P

* Lines 2..p+1:each of these Lines describes a legal operation. Line 2 describes the first operation, etc. Each line begins with a ' M ' for a move operation or a ' C ' for a count operation. For move operations, the line also contains the INTEGERS:X and y.for count operations, the line also contains a single in Teger:x.

Note that the value of N does not appear in the input file. No move operation would request a move a stack onto itself.

Output

Print the output from each of the count operations in the same order as the input file.

Sample Input

Sample Output

1 0 2

and check Set

 ?YesN (n<=30,000)Heap blocks, starting each heap is a
Box. Block number1–n.There are two kinds of actions:
?M x y: Indicates that the blockXWhere the heap is, pick it up and fold it
ToYOn the same heap.
? C x: Ask the  number of squares below the square x.
? The operation has a maximum of P (p<=100,000) times. For each C 
and output results.

Solution:
Apart fromParentArray, and also to open
SumArray: Records how many squares are in each heap.
IfParent[a] = A,Thea  block number. 
 under[i]  i How many 
           
1#include <iostream>
2#include <cstring>
3#include <cmath>
4usingnamespaceStd
5
6intfa[110000],sum[110000],under[110000];
7
8intGetfather (intA
9{
Tenif(fa[a]==a)
 OnereturnA
 AintT=getfather (Fa[a]);
 -Under[a]+=under[fa[a]];
 -fa[a]=t;
 thereturnT
 -}
 -
 -voidMergeintAintb
 +{
 -intTa=getfather (a);
 +intTb=getfather (b);
 Aif(TA==TB)
 at return;
 -UNDER[TA]=SUM[TB];
 -FA[TA]=TB;
 -SUM[TB]+=SUM[TA];
 -}
 -intMain ()
 in{
 -intN,a,b;
 toCharCh
 +cin>>n;
 - for(intI=1; i<=n;i++)
 the{
 *under[i]=0;
 $sum[i]=1;
Panax NotoginsengFa[i]=i;
 -}
 the while(n--)
 +{
 Acin>>ch;
 theif(ch=='M')
 +{
 -cin>>a>>b;
 $Merge (A, b);
 $}
 -Else
 -{
 thecin>>a;
 -Getfather (a);
Wuyicout <<under[a]<<endl;
 the}
 -}
 Wureturn0;
 -}View Code             
                

POJ 1988 Cube Stacking