POJ 3009--Curling 2.0【DFS】

Curling 2.0

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13329 Accepted: 5574

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.

0vacant square1block2start position3goal 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 13 26 61 0 0 2 1 01 1 0 0 0 00 0 0 0 0 30 0 0 0 0 01 0 0 0 0 10 1 1 1 1 16 11 1 2 1 1 36 11 0 2 1 1 312 12 0 1 1 1 1 1 1 1 1 1 313 12 0 1 1 1 1 1 1 1 1 1 1 30 0

Sample Output

14-1410-1

题目大意：

一种在宽为M高为N大小的矩阵上玩的冰壶游戏，起点字符为'2'，终点字符为'3'，矩阵上'0'为可移动区域，

'1'为石头区域。冰壶刚开始是静止的，每走一步都会选择某个方向运动，而且会沿着该方向一直运动不停，

也不会改变方向，除非冰壶碰到石头或者到达终点，才会停下(这算一步)。冰壶在运动的时候，不能改变方

向。冰壶碰到石头会变成静止状态，这时候石头会破裂，该区域变为可移动区域，而冰壶就可以改变方向了。

冰壶一旦走到终点就不再移动。问：冰壶从起点到终点最少停多少次(走多少步)？

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define maxn 25#define inf 0x3f3f3f3fusing namespace std;int w,h;int map[maxn][maxn];int sum;int d[4][2]={1,0,-1,0,0,1,0,-1};void dfs(int x, int y, int step){int i;if(step >= 10) return ;for (i = 0; i < 4; ++i){int fx = x + d[i][0];int fy = y + d[i][1];if(fx < 0 || fy < 0 || fx >=h || fy >=w)continue;if(map[fx][fy] == 1) //冰壶必须滑动时碰到石头，石头才会破裂。意味着若冰壶开始与石头相邻，则冰壶不能向这个方向滑动 continue;while(map[fx][fy]!=1 && map[fx][fy]!=3){fx = fx + d[i][0];fy = fy + d[i][1];if(fx < 0 || fy < 0 || fx >=h || fy >=w)break;}if(fx < 0 || fy < 0 || fx >=h || fy >=w)//这个判定也不能少。 continue ;if(map[fx][fy] == 1){//碰到石头 map[fx][fy] = 0;dfs(fx - d[i][0], fy - d[i][1], step + 1);//回到上一个位置，在从上一个位置开始深搜 map[fx][fy] = 1;}else if(map[fx][fy] == 3){sum = min(sum, step + 1);}}}int main (){while(scanf("%d%d", &w, &h)!= EOF){if(w == 0 && h == 0) break;int i, j;int sx, sy;for(i = 0; i < h; ++i)for(j = 0; j < w; ++j){scanf("%d", &map[i][j]);if(map[i][j] == 2){sx = i;sy = j;}}sum = inf;dfs(sx, sy, 0);//printf("%d\n", sum);if(sum != inf)printf("%d\n", sum);else printf("-1\n");}return 0;}