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2013
11-11

POJ 2585 Window Pains [解题报告] Java

Window Pains

问题描述 :

Boudreaux likes to multitask, especially when it comes to using his computer. Never satisfied with just running one application at a time, he usually runs nine applications, each in its own window. Due to limited screen real estate, he overlaps these windows and brings whatever window he currently needs to work with to the foreground. If his screen were a 4 x 4 grid of squares, each of Boudreaux’s windows would be represented by the following 2 x 2 windows:

11..
11..
....
....
.22.
.22.
....
....
..33
..33
....
....
....
44..
44..
....
....
.55.
.55.
....
....
..66
..66
....
....
....
77..
77..
....
....
.88.
.88.
....
....
..99
..99
When Boudreaux brings a window to the foreground, all of its squares come to the top, overlapping any squares it shares with other windows. For example, if window 1and then window 2 were brought to the foreground, the resulting representation would be:
122?
122?
????
????
If window 4 were then brought to the foreground:
122?
442?
44??
????
. . . and so on . . .

Unfortunately, Boudreaux’s computer is very unreliable and crashes often. He could easily tell if a crash occurred by looking at the windows and seeing a graphical representation that should not occur if windows were being brought to the foreground correctly. And this is where you come in . . .

输入:

Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets.

A single data set has 3 components:

  1. Start line – A single line:

    START

  2. Screen Shot – Four lines that represent the current graphical representation of the windows on Boudreaux’s screen. Each position in this 4 x 4 matrix will represent the current piece of window showing in each square. To make input easier, the list of numbers on each line will be delimited by a single space.
  3. End line – A single line:

    END

After the last data set, there will be a single line:

ENDOFINPUT

Note that each piece of visible window will appear only in screen areas where the window could appear when brought to the front. For instance, a 1 can only appear in the top left quadrant.

输出:

For each data set, there will be exactly one line of output. If there exists a sequence of bringing windows to the foreground that would result in the graphical representation of the windows on Boudreaux’s screen, the output will be a single line with the statement:

THESE WINDOWS ARE CLEAN

Otherwise, the output will be a single line with the statement:

THESE WINDOWS ARE BROKEN

样例输入:

START
1 2 3 3
4 5 6 6
7 8 9 9
7 8 9 9
END
START
1 1 3 3
4 1 3 3
7 7 9 9
7 7 9 9
END
ENDOFINPUT

样例输出:

THESE WINDOWS ARE CLEAN
THESE WINDOWS ARE BROKEN

解题代码:

//* @author: [email protected]
import java.io.*;
public class Main
{
 static int[][] map=new int[10][10];
 static int[][]	p=new int[4][4];
 public static void main(String[] args)throws IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  while(true)
  {
   String s=in.readLine();
   if(s.equals("ENDOFINPUT"))break;
   String[] ss;
   for(int i=0;i< 4;i++)
   {
	ss=in.readLine().split(" ");
	for(int j=0;j< 4;j++)
		p[i][j]=Integer.parseInt(ss[j]);
   }
   in.readLine();
   for(int i=1;i< 10;i++)
     for(int j=1;j< 10;j++)
         map[i][j]=0;
   f();
   for(int i=1;i< 10;i++)
	map[i][i]=0;		
   while(true)
   {
      boolean qq=false;
      for(int i=1;i< 10;i++)
	for(int j=1;j< 10;j++)
          for(int k=1;k< 10;k++)
	    if(map[i][j]==1&&map[j][k]==1&&map[i][k]==0)
	    {
		map[i][k]=1;
		qq=true;
	    }
      if(!qq)break;
    }
    boolean bb=true;
    for(int i=1;i< 10;i++)
	if(map[i][i]==1){
          bb=false;
	  break;
	}
    if(bb) System.out.println("THESE WINDOWS ARE CLEAN");
    else System.out.println("THESE WINDOWS ARE BROKEN");
   }
 }

  static void f()
  {
        int x;
 	x=p[0][1];
	map[x][1]=1;
	map[x][2]=1;
	x=p[0][2];
	map[x][2]=1;
	map[x][3]=1;
	x=p[1][0];
	map[x][1]=1;
	map[x][4]=1;
	x=p[2][0];
	map[x][4]=1;
	map[x][7]=1;
	x=p[1][3];
	map[x][3]=1;
	map[x][6]=1;
	x=p[2][3];
	map[x][6]=1;
	map[x][9]=1;
	x=p[3][1];
	map[x][7]=1;
	map[x][8]=1;
	x=p[3][2];
	map[x][8]=1;
	map[x][9]=1;
	x=p[1][1];
	map[x][1]=1;
	map[x][2]=1;
	map[x][4]=1;
	map[x][5]=1;
	x=p[1][2];
	map[x][2]=1;
	map[x][3]=1;
	map[x][5]=1;
	map[x][6]=1;
	x=p[2][1];
	map[x][4]=1;
	map[x][5]=1;
	map[x][7]=1;
	map[x][8]=1;
	x=p[2][2];
	map[x][5]=1;
	map[x][6]=1;
	map[x][8]=1;
	map[x][9]=1;
   }
}

  1. 有一点问题。。后面动态规划的程序中
    int dp[n+1][W+1];
    会报错 提示表达式必须含有常量值。该怎么修改呢。。