poj 3009 Curling 2.0

                                                                                                      Curling 2.0

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18430		Accepted: 7573

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：扔球，把图中某一点作为球的起始点，之后球可以从上下左右四个方向扔（若障碍物就在要扔的方向的前面一格，则不能往这个方向扔），往前仍之后，若碰到障碍物，则球停止运动，且碰到的障碍物将消失。重复以上步骤，找到一种策略，使得球到达终点，并且使得扔球的次数达到最少，输出该最小次数，若找不到这种策略或者达到终点所用次数超过10次，输出-1.
思路：深度优先搜索，搜索最小的扔球次数。
注意：题目行列是读反的。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
const int N_MAX = 30;
int field[N_MAX][N_MAX];
int W, H;
const int direction[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
int min_step;
void dfs(const int&x,const int&y,int step) {
    if (step > 10 || step > min_step)return;//若当前的情况走的step比已知的min_step还大，这种情况就直接排除了
    for (int i = 0;i < 4;i++) {
        int dx = x;
        int dy = y;
        while(dx >= 0 && dx < W&&dy >= 0 && dy < H) {
            switch (field[dx][dy]) {
            case 0: 
                dx +=direction[i][0];
                dy +=direction[i][1];
                break;
            case 3: 
                if (step + 1 < min_step)//找到一种新的情况，马上比较，能否能找到更小的步骤数
                    min_step = step+1;
                dx = -1;
                break;//dx=-1是为了退出while循环
            case 1: //遇到障碍物
                if (!(dx - direction[i][0] == x&&dy - direction[i][1] == y)) {//若球还没运动，面向的方向是墙壁，就直接换方向
                   field[dx][dy] = 0;///////////////
                    dfs(dx-direction[i][0], dy-direction[i][1], step + 1);
                    field[dx][dy] = 1;//复原
                }
               dx = -1;
               break;
            default:
                break;
            }
        }
    }
}
int main() {
    while (scanf("%d%d", &W, &H)&&W) {
        for (int i = 0;i < H;i++) {//////////////////////
            for (int j = 0;j < W;j++) {
                scanf("%d",&field[j][i]);
            }
        }
        int x, y;
        for (int i = 0;i < H;i++) {
            for (int j = 0;j < W;j++) {
                if (field[j][i] == 2) {
                    x = j;y = i;////////////////////////
                    break;
                }
            }
        }
        field[x][y] = 0;
     min_step = 11;
        dfs(x, y, 0);
        if (min_step > 10)printf("-1
");
        else printf("%d
",min_step);
    }
    return 0;
}