POJ 3009 Curling 2.0

Curling 2.0

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12556		Accepted: 5291

Description

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

Input

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

Output

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.

Sample Input

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

Sample Output

1
4
-1
4
10
-1

题意：求冰壶从S到G的最短路径，但与一般的搜索问题不同的是，冰壶每次走的步数都不一定，因为冰壶走起来不会停，除非碰到石头或走到了G的位置。若碰到石头的话，石头会消失，冰壶停在石头消失前的位置的邻近一格。需要注意的是，冰壶若越界，则游戏结束。并且走的步数不能超过10。

因为每走一步石头会消失导致情况都不一样，而且若没到达G，返回上一层的时候石头还要恢复，因此不能用BFS

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>

using namespace std;

int Map[30][30];
int Min;
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int h,w;

int judge(int x,int y)//判断冰壶是否越界
{
	if(x>0&&x<=h&&y>0&&y<=w)
		return 1;
	return 0;
}

int DFS(int x,int y,int step)
{
	if(step>10)//剪枝
		return 0;
	int nx,ny;
	step++;//记录步数
	for(int i=0;i<4;i++)
	{
		nx=x+dx[i];
		ny=y+dy[i];
		if(Map[nx][ny]!=1&&judge(nx,ny))//冰壶没有越界，且冰壶走的方向的下一格没有石头
		{
			while(Map[nx][ny]!=3&&Map[nx][ny]!=1&&judge(nx,ny))
			{
				nx+=dx[i];
				ny+=dy[i];
			}
			if(Map[nx][ny]==1)//冰壶撞到石头
			{
				Map[nx][ny]=0;//冰壶撞到后，石头消失
				DFS(nx-dx[i],ny-dy[i],step);//冰壶停在石头的前一格上,向下遍历
				Map[nx][ny]=1;//恢复石头
			}
			else if(Map[nx][ny]==3)//冰壶到达出口
			{
				if(Min>step)
					Min=step;
				//printf("%d\n",step);
				return 1;
			}
		}
	}
}

int main()
{
	//int w,h;
	int x,y;
	while(scanf("%d%d",&w,&h)!=EOF)
	{
		if(w==0&&h==0)
			break;
		memset(Map,0,sizeof(Map));
		for(int i=1;i<=h;i++)
		{
			for(int j=1;j<=w;j++)
			{
				scanf("%d",&Map[i][j]);
				if(Map[i][j]==2)//记录入口位置
				{
					x=i;
					y=j;
				}
			}
		}
		Min=999;
		DFS(x,y,0);
		if(Min<=10)
			printf("%d\n",Min);
		else
			printf("-1\n");
	}
	return 0;
}