Poj 1988 cube stacking (and query set)

Cube stacking

Time limit:2000 ms		Memory limit:30000 K
Total submissions:14901		Accepted:5037
Case time limit:1000 ms

Description

Farmer John and Betsy are 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 are two 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
Number of cubes on the stack with cube X that are under the cube 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 two integers: X and Y. For Count
Operations, the line also contains a single INTEGER: X.

Note that the value for N does not appear in the input file. No move operation will 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

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

Sample output

 
102
 

Source

 

In the hyper-era, although the function is implemented, it is always time-out, because there is no real grasp and query set usage method:

 /*  * Method 1: timeout. Here a lower is used to locate the next one.CubeBlock, * Here we use a loop, so we will traverse the lower cube every time, resulting in timeout.  */  # Include <Iostream># Include <Stdio. h> # Define Maxnum30001 Using   Namespace  STD; typedef  Struct  {  Int Count; //  Record the number of all cubes below the cube      Int Parent; //  Point to the cube at the top of the heap where the cube is located      Int Lower; // The block pointing to the bottom of the current cube  } Cube; cube [maxnum];  Void  Init (){  For ( Int I = 1 ; I <maxnum; I ++ ) {Cube [I]. Count = 0 ; //  Initialize 0 at the bottom of each cube Cube [I]. Parent = I; //  Initialize each cube as a single heap and the top of the current heap Cube [I]. Lower =0  ;}}  //  Merge the heap of X and Y  Void Union_set ( Int X, Int  Y ){  Int Lower_parent; //  The parent node of the grid in the heap below      Int Lower_cube; //  Lower lattice subscript      Int Parent;//  Parent node subscript of the top heap  Parent = Cube [X]. parent; lower_parent = Cube [Y]. parent; lower_cube = X;  //  Update the stack above. All the Count values under the current cube must change.      While (Cube [lower_cube]. Lower! = 0  ) {Cube [lower_cube]. Count = Cube [lower_parent]. Count + 1  ; Lower_cube =Cube [lower_cube]. Lower;} cube [lower_cube]. Count = Cube [lower_parent]. Count + 1  ;  //  At the same time, the bottom of the bottom block is updated, that is, the top block of the bottom stack. Cube [lower_cube]. Lower = Lower_parent;  //  Update the parent nodes of all the blocks below, which becomes a large heap      While (Cube [lower_parent]. Lower! = 0  ) {Cube [lower_parent]. Parent = Parent; lower_parent =Cube [lower_parent]. Lower;} cube [lower_parent]. Parent = Parent ;}  Int  Main (){  //  There are P operations      Int  P;  //  Command type,  //  M x Y: Move the heap where X cube is located to the heap where y cube is located.  //  C x outputs the number of cubes under X on the heap where X is located.      Char Ch;  Int  X, Y; Init (); scanf (  "  % D  " ,& P );  While (P -- ) {Scanf (  "  \ N % C  " ,& Ch );  //  Getchar ();  // Cin> CH;          If (CH = '  M  '  ) {Scanf (  "  % D  " , & X ,& Y); union_set (x, y );  //  Merge set  }  Else   If (CH = ' C  '  ) {Scanf (  "  % D  " ,& X); printf (  "  % D \ n  "  , Cube [X]. Count );}}  Return   0  ;} 

Query the ACCode

# Include <iostream># Include <Stdio. h> # Define Maxnum30010 Using   Namespace  STD; typedef  Struct  {  Int Count; //  Record the number of all cubes below the cube      Int Parent; //  Point to the cube at the top of the heap where the cube is located      Int  Num;} cube; cube [maxnum]; Void  Init (){  For ( Int I = 1 ; I <maxnum; I ++ ) {Cube [I]. Count = 0 ; //  Initialize 0 at the bottom of each cube Cube [I]. num = 1  ; Cube [I]. Parent =- 1 ; // Initializes each cube as a single heap and is at the bottom of the current heap.  }}  Int Find ( Int  A ){  //  Point to yourself, indicating that you are the root      If (Cube [A]. Parent =- 1  )  Return  A;  Int TEM = Cube [A]. parent; cube [A]. Parent = Find (cube [A]. Parent ); // The compression path, so that all the roots are the same Cube [A]. Count + = cube [TEM]. count; //  Update the number below the current block, which is      Return Cube [A]. parent; //  Returns the ultimate root.  }  Void Union_set ( Int X, Int  Y) {cube [X]. Parent = Y; cube [X]. Count + = Cube [Y]. Num; cube [Y]. Num + =Cube [X]. Num ;}  Int  Main (){  //  There are P operations      Int  P;  //  Command type,  //  M x Y: Move the heap where X cube is located to the heap where y cube is located.  //  C x outputs the number of cubes under X on the heap where X is located.      Char  Ch;  Int X, Y, RX, Ry; Init (); scanf (  "  % D  " ,& P );  While (P -- ){  //  Scanf ("% s", CH ); Cin> Ch;  If (CH = '  M  '  ) {Scanf ( "  % D  " , & X ,& Y); RX = Find (x); ry = Find (y );  If (RX! = Ry) union_set (RX, ry );  //  Merge set  }  Else   If (CH = '  C '  ) {Scanf (  "  % D  " ,& X); find (x); printf (  "  % D \ n  "  , Cube [X]. Count );}}  Return   0  ;}