Question

我想在运行程序时自动解决迷宫问题。

一开始我的迷宫就是这样。

最后应该看起来像这样：

我有一个3维的数组（用于行，列和侧面）。

侧面可以在（0），右（1），上方（2）和左（3）处。因此，我检查每个单元是否有墙。如果是，我在该单元格中输入是。

mazeTab[0, 0, 0] = 0;
mazeTab[0, 0, 1] = 1;
mazeTab[0, 0, 2] = 1;
mazeTab[0, 0, 3] = 1;

mazeTab[1, 0, 0] = 0;
mazeTab[1, 0, 1] = 0;
mazeTab[1, 0, 2] = 1;
mazeTab[1, 0, 3] = 1;

mazeTab[2, 0, 0] = 0;
mazeTab[2, 0, 1] = 0;
mazeTab[2, 0, 2] = 1;
mazeTab[2, 0, 3] = 0;

mazeTab[3, 0, 0] = 0;
mazeTab[3, 0, 1] = 1;
mazeTab[3, 0, 2] = 1;
mazeTab[3, 0, 3] = 0;
//=================================

mazeTab[0, 1, 0] = 0;
mazeTab[0, 1, 1] = 0;
mazeTab[0, 1, 2] = 1;
mazeTab[0, 1, 3] = 1;

mazeTab[1, 1, 0] = 1;
mazeTab[1, 1, 1] = 1;
mazeTab[1, 1, 2] = 0;
mazeTab[1, 1, 3] = 0;

mazeTab[2, 1, 0] = 1;
mazeTab[2, 1, 1] = 1;
mazeTab[2, 1, 2] = 1;
mazeTab[2, 1, 3] = 1;

mazeTab[3, 1, 0] = 0;
mazeTab[3, 1, 1] = 1;
mazeTab[3, 1, 2] = 0;
mazeTab[3, 1, 3] = 1;
//===================================
mazeTab[0, 2, 0] = 0;
mazeTab[0, 2, 1] = 1;
mazeTab[0, 2, 2] = 0;
mazeTab[0, 2, 3] = 1;

mazeTab[1, 2, 0] = 1;
mazeTab[1, 2, 1] = 1;
mazeTab[1, 2, 2] = 1;
mazeTab[1, 2, 3] = 1;

mazeTab[2, 2, 0] = 1;
mazeTab[2, 2, 1] = 1;
mazeTab[2, 2, 2] = 1;
mazeTab[2, 2, 3] = 1;

mazeTab[3, 2, 0] = 0;
mazeTab[3, 2, 1] = 1;
mazeTab[3, 2, 2] = 0;
mazeTab[3, 2, 3] = 1;
//===================================
mazeTab[0, 3, 0] = 1;
mazeTab[0, 3, 1] = 1;
mazeTab[0, 3, 2] = 0;
mazeTab[0, 3, 3] = 1;

mazeTab[1, 3, 0] = 1;
mazeTab[1, 3, 1] = 1;
mazeTab[1, 3, 2] = 1;
mazeTab[1, 3, 3] = 1;

mazeTab[2, 3, 0] = 1;
mazeTab[2, 3, 1] = 1;
mazeTab[2, 3, 2] = 1;
mazeTab[2, 3, 3] = 1;

mazeTab[3, 3, 0] = 0;
mazeTab[3, 3, 1] = 1;
mazeTab[3, 3, 2] = 0;
mazeTab[3, 3, 3] = 1;

然后我使用一种方法来解决整个迷宫，但是我很确定某些地方出了问题。使用它时，我检查是否有下一个单元格，例如：如果我有一个0且我可以走到上方，我检查上方的单元格是否也可以转到下方的单元格，以验证我们是否没有墙在那里。

   int solvemaze(int horizontal, int vertical,int cote)
    {

        //horizontalestination is the last cell(maze[taille-1][taille-1])
        if ((horizontal == taille - 1) && (vertical == taille - 1))
        {
            solution[horizontal,vertical] = 1;
            return 1;
        }

        if (horizontal >= 0 && vertical >= 0 && horizontal < taille && vertical < taille && solution[horizontal,vertical] == 0)
        {

            printsolution();
            //if safe to visit then visit the cell
            solution[horizontal,vertical] = 1;
            //under
            if (mazeTab[horizontal,vertical,cote]==0)
                return solvemaze(horizontal,vertical+1,(cote+1)%4);
            cote = (cote + 1) % 4;
            //right
            if (mazeTab[horizontal, vertical, 1] == 0)
                return solvemaze(horizontal+1, vertical, (cote + 1)% 4);
            cote = (cote + 1) % 4;
            //up
            if (mazeTab[horizontal, vertical, 2] == 0)
                return solvemaze(horizontal, vertical -1, (cote + 1) % 4);
            cote = (cote + 1) % 4;
            //left
            if (mazeTab[horizontal, vertical, 3] == 0)
                return solvemaze(horizontal -1, vertical, (cote + 1) % 4);
            cote = (cote + 1) % 4;
            //backtracking
            solution[horizontal,vertical] = 0;
            return 1;
        }
        return 1;

提醒您，如果您可以转到特定的单元格，则为1。因此解决迷宫的路径充满了1。

最后，我从0,3,0单元格开始迷宫，也就是这个迷宫：

1 0 0 0
0 0 1 0
0 1 1 0
**0** 1 1 0

当我跑步时，我得到的是这样：

以下0应该上交1。

0 1 1 1 
1 1 0 **0** 
1 0 0 **0** 
**0** 0 0 **0**

你们能帮我找到错误吗？

Answer 1

由于@HHLV，我已经弄清楚了如何解决此问题。

所以我的方法resolvemaze看起来像这样：

{{1}}

知道它可以工作，谢谢！

解决迷宫C＃

1 个答案: