POJ 3009 Curling 2.0 （dfs）

POJ 3009 Curling 2.0 （dfs）

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)<喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"center">
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

題意：一種游戲，規則是：從起點S出發，在沒有障礙的直線上會一直不停的行進直到遇到障礙或到達終點G停止。在遇到障礙時，要停在障礙前，然後障礙就會消除，然後重新選擇四個方向行進。輸入圖中2代表起點，3代表終點，0代表無障礙區域，1代表障礙區域。求要從起點到達終點最少需要停多少次。

解析：dfs。

PS：C++交的結果是RE，G++是AC

AC代碼：

#include
#include
#include
#include
using namespace std;
int vis[25][25];
int n,m;
int sx,sy,ex,ey;
int flag,minstep;        //flag標記是否找到路徑,minstep記錄總的步數
int d[4][2] = {1,0,0,1,-1,0,0,-1};   //移動的方向

void dfs(int x, int y, int step){    //當前位置，步數
    int dx, dy;
    if(step >= 10) return ;
    for(int i=0; i<4; i++){
        if(!vis[ x+d[i][0] ][ y+d[i][1] ]){
            dx = x;
            dy = y;
            while(!vis[ dx+d[i][0] ][ dy+d[i][1] ]){    //判斷下一個節點是否是障礙
                dx = dx + d[i][0];                      //符合條件才加
                dy = dy + d[i][1];
                if(dx == ex && dy == ey){
                    if(minstep > step + 1) minstep = step + 1;
                    flag = 1;
                    return ;
                }
                if(dx < 0 || dx >= n || dy < 0 || dy >= m) break;  //判斷越界
            }
            if(dx >= 0 && dx < n && dy >= 0 && dy < m && step + 1 < 10){    //與障礙撞上後
                vis[ dx+d[i][0] ][ dy+d[i][1] ] = 0;      //去除障礙
                dfs(dx, dy, step+1);
                vis[ dx+d[i][0] ][ dy+d[i][1] ] = 1;      //回溯
            }  
        }
    }
}

int main()
{
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk


    int i,j;
    while(scanf("%d%d", &m, &n)!=EOF && !(!m && !n)){
        memset(vis, 0, sizeof(vis));              //不加這句，竟然WA
        for(i=0; i