poj 3009 Curling 2.0（dfs）

Language: Default Curling 2.0 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11926   Accepted: 5051 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 Source Japan 2006 Domestic

题意：冰球游戏，遇到障碍物（1）就可以撞碎，求从2到3的最小步数

注意点：

   1、    如果这个点事起点，周围的点有障碍物，那么不能撞，即不能走这个方向（第三个样例）

  2、   最多10次移动冰球，剪枝

  3、 撞碎障碍物后，停在障碍物前面

其他的在代码中：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)
using namespace std;
#define N 30
#define INF 0x3f3f3f3f

int a[N][N],ans;
int n,m;
int step[4][2]={{0,-1},{1,0},{0,1},{-1,0}};

int judge(int x,int y)
{
	 if(x>=0&&x<n&&y>=0&&y<m)
		return 1;
	 return 0;
}

void dfs(int x,int y,int pos)
{
	if(pos>10) return ;

	if(a[x][y]==3)
	{
		if(ans>pos)
			ans=pos;
		return ;
	}

	int i;
	for(i=0;i<4;i++)
	{
		int xx=x;
		int yy=y;
         if(a[xx+step[i][0]][yy+step[i][1]]==1)          		//冰球还没有启动时如果这个方向有1，那么不能撞破这个1
			continue;

		while(1)
		{
			xx=xx+step[i][0];
			yy=yy+step[i][1];

			if(!judge(xx,yy))
				break;

			if(a[xx][yy]==1)
			{
				a[xx][yy]=0;
				dfs(xx-step[i][0],yy-step[i][1],pos+1);  //撞碎障碍物后，停在障碍物前面
				a[xx][yy]=1;
				break;
			}
			if(a[xx][yy]==3)
				dfs(xx,yy,pos);
	   }
	}
}

int main()
{
	int i,j,sx,sy;
	while(~scanf("%d%d",&m,&n),n+m)
	{
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
			{
				scanf("%d",&a[i][j]);
			     if(a[i][j]==2)
				 {
				 	sx=i;
				 	sy=j;
				 	a[sx][sy]=0;
				 }
			}
		 ans=INF;
		 dfs(sx,sy,1);
		 if(ans==INF)
			printf("-1\n");
		 else
			printf("%d\n",ans);
	}
   return 0;
}