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2013
11-12

POJ 3009 Curling 2.0 [解题报告] Java

Curling 2.0

问题描述 :

On Planet MM-21, after their Olympic games this year, curling is getting popular. But the rules are somewhat different from ours. The game is played on an ice game board on which a square mesh is marked. They use only a single stone. The purpose of the game is to lead the stone from the start to the goal with the minimum number of moves.

Fig. 1 shows an example of a game board. Some squares may be occupied with blocks. There are two special squares namely the start and the goal, which are not occupied with blocks. (These two squares are distinct.) Once the stone begins to move, it will proceed until it hits a block. In order to bring the stone to the goal, you may have to stop the stone by hitting it against a block, and throw again.


Fig. 1: Example of board (S: start, G: goal)

The movement of the stone obeys the following rules:

  • At the beginning, the stone stands still at the start square.
  • The movements of the stone are restricted to x and y directions. Diagonal moves are prohibited.
  • When the stone stands still, you can make it moving by throwing it. You may throw it to any direction unless it is blocked immediately(Fig. 2(a)).
  • Once thrown, the stone keeps moving to the same direction until one of the following occurs:
    • The stone hits a block (Fig. 2(b), (c)).
      • The stone stops at the square next to the block it hit.
      • The block disappears.
    • The stone gets out of the board.
      • The game ends in failure.
    • The stone reaches the goal square.
      • The stone stops there and the game ends in success.
  • You cannot throw the stone more than 10 times in a game. If the stone does not reach the goal in 10 moves, the game ends in failure.


Fig. 2: Stone movements

Under the rules, we would like to know whether the stone at the start can reach the goal and, if yes, the minimum number of moves required.

With the initial configuration shown in Fig. 1, 4 moves are required to bring the stone from the start to the goal. The route is shown in Fig. 3(a). Notice when the stone reaches the goal, the board configuration has changed as in Fig. 3(b).


Fig. 3: The solution for Fig. D-1 and the final board configuration

输入:

The input is a sequence of datasets. The end of the input is indicated by a line containing two zeros separated by a space. The number of datasets never exceeds 100.

Each dataset is formatted as follows.

the width(=w) and the height(=h) of the board
First row of the board

h-th row of the board

The width and the height of the board satisfy: 2 <= w <= 20, 1 <= h <= 20.

Each line consists of w decimal numbers delimited by a space. The number describes the status of the corresponding square.

0 vacant square
1 block
2 start position
3 goal position

The dataset for Fig. D-1 is as follows:

6 6
1 0 0 2 1 0
1 1 0 0 0 0
0 0 0 0 0 3
0 0 0 0 0 0
1 0 0 0 0 1
0 1 1 1 1 1

输出:

For each dataset, print a line having a decimal integer indicating the minimum number of moves along a route from the start to the goal. If there are no such routes, print -1 instead. Each line should not have any character other than this number.

样例输入:

2 1
3 2
6 6
1 0 0 2 1 0
1 1 0 0 0 0
0 0 0 0 0 3
0 0 0 0 0 0
1 0 0 0 0 1
0 1 1 1 1 1
6 1
1 1 2 1 1 3
6 1
1 0 2 1 1 3
12 1
2 0 1 1 1 1 1 1 1 1 1 3
13 1
2 0 1 1 1 1 1 1 1 1 1 1 3
0 0

样例输出:

1
4
-1
4
10
-1

解题代码:

//* @author 

import java.util.*;
public class Main {

static final int MAX= 30;
static int w,h;
static int start_x,start_y;
static int flag[][] = new int[MAX][MAX];
static int min=11;
static int step=1;

static void dfs(int x,int y,int step)
{
	if(step>10) return;
	if(step==min) return;
for(int i=x+1;i< h;i++)
{
 if(i==x+1&&flag[i][y]==1) break;
 if(flag[i][y]==3){if(min>step) min=step;return;}
 if(flag[i][y]==1&&step< min){flag[i][y]=0;dfs(i-1,y,step+1);flag[i][y]=1;break;}
}
for(int i=x-1;i>=0;i--)
{
 if(i==x-1&&flag[i][y]==1) break;
 if(flag[i][y]==3){if(min>step);min=step;return;}
 if(flag[i][y]==1&&step< min){flag[i][y]=0;dfs(i+1,y,step+1);flag[i][y]=1;break;}
 }
for(int i=y+1;i< w;i++)
{
 if(i==y+1&&flag[x][i]==1) break;
 if(flag[x][i]==3){if(min>step) min=step;return;}
 if(flag[x][i]==1&&step< min){flag[x][i]=0;dfs(x,i-1,step+1);flag[x][i]=1;break;}
}
for(int i=y-1;i>=0;i--)
{
 if(i==y-1&&flag[x][i]==1) break;
 if(flag[x][i]==3&&step< min){if(min>step) min=step;return;}
 if(flag[x][i]==1&&step< min){flag[x][i]=0;dfs(x,i+1,step+1);flag[x][i]=1;break;}
}
return;
}

public static void main(String[] args) {
 Scanner in = new Scanner(System.in);	
  while(true)
  {
	w=in.nextInt();
	h=in.nextInt();
	if(w==0&&h==0) break;
	for(int i=0;i< h;i++)
	{
          for(int j=0;j< w;j++)
	   {
		flag[i][j]=in.nextInt();
		if(flag[i][j]==2) 
		{
	        start_x=i;
		 start_y=j;
		 flag[i][j]=0;
	       }
	     }
	}
	step=1;
	min=11;
	dfs(start_x,start_y,step);
	if(min==11) System.out.println("-1");
	else System.out.println(min); 
  }
 }
}

  1. 您没有考虑 树的根节点是负数的情况, 若树的根节点是个很大的负数,那么就要考虑过不过另外一边子树了

  2. Thanks for taking the time to examine this, I really feel strongly about it and love studying a lot more on this topic. If possible, as you acquire experience

  3. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术,也不懂什么是算法,从而也不要求程序员懂什么算法,做程序从来不考虑性能问题,只要页面能显示出来就是好程序,这是国内的现状,很无奈。