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2013
11-10

POJ 1988 Cube Stacking [解题报告] Java

Cube Stacking

问题描述 :

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 the 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.

输入:

* 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.

输出:

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

样例输入:

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

样例输出:

1
0
2

解题代码:

/* @author: */
import java.util.Scanner;
import java.util.Arrays;
public class Main{
 static int p[]=new int[30010];
 static int d[]=new int[30010];
 static int num[]=new int[30010];
 static int i=0;
 static char c[]=new char[5];

 static int getup(int a)
 {
  int b=a;
   while(p[b]!=b)
    b=p[b];
  return p[a]=b;
 }

 static int getd(int a)
 {
   int b=a;
   int v=0;
   while(d[b]!=b)
   {
    v+=num[b];
    b=d[b];
   }
   num[a]=v;
   return d[a]=b;
 }

 public static void main(String[] args){   
  Scanner sc=new Scanner(System.in);
  for(i=1;i< 30005;i++)
   p[i]=d[i]=i;
  Arrays.fill(num,0);
  int n,m,u,v;
  n=sc.nextInt();
  while((n--)!=0)
  {
	c=sc.next().toCharArray();
	if(c[0]=='C')
	{
         m=sc.nextInt();
	  getd(m);
	  System.out.printf("%d\n",num[m]);
	}
	else if(c[0]=='M')
	{
	  u=sc.nextInt();
         v=sc.nextInt();
	  int g1=getd(u);
	  int g2=getup(v);
	  d[g1]=g2;
	  p[g2]=g1;
	  num[g1]=1;
	}
   }
  }
}