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2013
11-09

POJ 1077 Eight [解题报告] Java

Eight

问题描述 :

The 15-puzzle has been around for over 100 years; even if you don’t know it by that name, you’ve seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let’s call the missing tile ‘x’; the object of the puzzle is to arrange the tiles so that they are ordered as:
 1  2  3  4 

5 6 7 8
9 10 11 12
13 14 15 x

where the only legal operation is to exchange ‘x’ with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:

 1  2  3  4    1  2  3  4    1  2  3  4    1  2  3  4 

5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->

The letters in the previous row indicate which neighbor of the ‘x’ tile is swapped with the ‘x’ tile at each step; legal values are ‘r’,'l’,'u’ and ‘d’, for right, left, up, and down, respectively.

Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and

frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing ‘x’ tile, of course).

In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three

arrangement.

输入:

You will receive a description of a configuration of the 8 puzzle. The description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus ‘x’. For example, this puzzle
 1  2  3 

x 4 6
7 5 8

is described by this list:


1 2 3 x 4 6 7 5 8

输出:

You will print to standard output either the word “unsolvable”, if the puzzle has no solution, or a string consisting entirely of the letters ‘r’, ‘l’, ‘u’ and ‘d’ that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line.

样例输入:

 2  3  4  1  5  x  7  6  8 

样例输出:

ullddrurdllurdruldr

解题代码:

//* @author: [email protected]
import java.util.*;
public class Main
{
	static int[][] arr;
	static boolean[] bb=new boolean[10000000];
	static Queue< my> qu=new LinkedList< my>();
	public static void main(String[] args)
	{
		Scanner in=new Scanner(System.in);
		arr=new int[5][5];
		String s;
		for(int i=1;i< 4;i++)
		{
			for(int j=1;j< 4;j++)
			{
				s=in.next();
				if(s.equals("x"))arr[i][j]=0;
				else arr[i][j]=Integer.parseInt(s);
			}
		}
		int u=getNum();
		bfs(u);
		
	}
	static void bfs(int t)
	{
			qu.add(new my("",t));
		while(!qu.isEmpty())
		{
			my h=qu.poll();
			int u=h.u;
			String s=h.s;
			if(u==123456780)
			{
				System.out.println(s);
				return;
			}
			if(bb[u%9999991])continue;
			bb[u%9999991]=true;
			int i=-1,j=-1,p=u;
			for(int u1=3;u1>0;u1--)
			{
				for(int u2=3;u2>0;u2--)
				{
					arr[u1][u2]=p%10;
					if(arr[u1][u2]==0)
					{
						i=u1;
						j=u2;
					}
					p/=10;
				}
			}
			change(i,j,i-1,j);
			int y=getNum();
			qu.add(new my(s+"u",y));
			change(i-1,j,i,j);
			
			change(i,j,i+1,j);
			y=getNum();
			qu.add(new my(s+"d",y));
			change(i+1,j,i,j);
			
			change(i,j,i,j+1);
			y=getNum();
			qu.add(new my(s+"r",y));
			change(i,j+1,i,j);
			
			change(i,j,i,j-1);
			y=getNum();
			qu.add(new my(s+"l",y));
			change(i,j-1,i,j);
		}
		System.out.println("unsolvable");
	}
	static int getNum()
	{
		int t=0;
		for(int i=1;i< 4;i++)
			for(int j=1;j< 4;j++)
			{
				t*=10;
				t+=arr[i][j];
			}
		return t;
	}
	static void change(int x1,int y1,int x2,int y2)
	{
		arr[x1][y1]=arr[x2][y2];
		arr[x2][y2]=0;
	}
}
class my
{
	String s="";
	int u;
	public my(String t,int a)
	{
		u=a;
		s=t;
	}
}

  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. for(int i=1; i<=m; i++){
    for(int j=1; j<=n; j++){
    dp = dp [j-1] + 1;
    if(s1.charAt(i-1) == s3.charAt(i+j-1))
    dp = dp[i-1] + 1;
    if(s2.charAt(j-1) == s3.charAt(i+j-1))
    dp = Math.max(dp [j - 1] + 1, dp );
    }
    }
    这里的代码似乎有点问题? dp(i)(j) = dp(i)(j-1) + 1;这个例子System.out.println(ils.isInterleave("aa","dbbca", "aadbbcb"));返回的应该是false