所以，基本上，我正在尝试编写一个解决迷宫的程序。我对不同迷宫进行了多次测试，我发现我的程序无法解决所有类型的迷宫，只能解决少数几种，因为有些特定的死胡同会让我的程序卡住，无法返回。
我的代码背后的逻辑基本上是遍历整个迷宫直到找到出口，如果在这一过程中遇到死胡同，它应该能够返回并找到一条新的未探索路径。
我的代码一直运行良好，直到我开始用更复杂的迷宫进行测试，这些迷宫包含不同类型的棘手死胡同。例如：
maze = ( [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], [1,0,1,1,1,0,1,1,1,1,1,1,1,1,1], [1,0,0,0,0,0,0,0,1,0,0,0,0,0,1], [1,1,1,0,1,1,1,0,1,0,1,0,1,0,1], [1,1,1,0,1,0,1,0,1,0,1,1,1,0,1], [1,1,1,1,1,0,1,0,1,0,0,1,0,0,1], [1,1,0,0,0,0,1,0,1,1,0,1,0,1,1], [1,1,0,1,1,0,0,0,0,1,0,1,0,1,1], [1,1,0,0,1,1,0,1,0,1,0,1,0,0,1], [1,1,0,1,0,0,1,0,0,0,1,0,0,1,1], [1,1,1,1,1,1,1,1,1,1,1,3,1,1,1] )这是一个我的程序可以解决的迷宫示例，其中1表示墙壁，0表示自由路径，3表示终点，起点是maze[1][1]
maze = ( [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], [1,0,0,0,0,0,0,0,0,0,0,0,1,1,1], [1,0,1,1,1,0,1,1,1,1,1,1,0,1,1], [1,0,1,0,0,0,0,0,0,1,1,1,0,1,1], [1,0,0,0,1,1,1,1,0,0,0,1,0,0,1], [1,1,0,1,1,0,0,1,1,1,0,1,1,0,1], [1,1,0,1,1,1,0,0,0,1,0,1,0,0,1], [1,0,0,1,1,1,1,1,0,1,0,1,1,0,1], [1,0,1,1,0,0,0,0,0,1,0,0,0,0,1], [1,0,0,0,0,1,1,1,1,1,0,1,1,1,1], [1,1,1,1,1,1,1,1,1,1,3,1,1,1,1] )现在是一个我的程序无法解决的迷宫。我认为这个迷宫的问题在于它有”L”形或类似锯齿形的死胡同，但说实话，我不知道发生了什么。例如，maze[5][5]处的死胡同（我的程序在这里卡住）
在展示代码之前，我想解释一些关于它的主题：
- 我打印的所有字符串都是英文和葡萄牙语，语言之间用”/”分隔，所以不用担心这一点。
- 为了探索迷宫，程序使用递归策略，并用2标记已探索的路径。
- 程序使用基于数组的x和y坐标。x + 1向下移动，x – 1向上移动，y + 1向右移动，y – 1向左移动。
- 如果它在死胡同中卡住，策略是查看周围的情况以发现这是哪种类型的死胡同，然后返回直到找到标记为0的新路径。
这些是你将在我代码中看到的死胡同情况。
- 如果可能的话，我想得到一些关于如何改进我的代码或使其更清晰的建议。
那么，我们开始吧：
maze = ( [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], [1,0,0,0,0,0,0,0,0,0,0,0,1,1,1], [1,0,1,1,1,0,1,1,1,1,1,1,0,1,1], [1,0,1,0,0,0,0,0,0,1,1,1,0,1,1], [1,0,0,0,1,1,1,1,0,0,0,1,0,0,1], [1,1,0,1,1,0,0,1,1,1,0,1,1,0,1], [1,1,0,1,1,1,0,0,0,1,0,1,0,0,1], [1,0,0,1,1,1,1,1,0,1,0,1,1,0,1], [1,0,1,1,0,0,0,0,0,1,0,0,0,0,1], [1,0,0,0,0,1,1,1,1,1,0,1,1,1,1], [1,1,1,1,1,1,1,1,1,1,3,1,1,1,1] )def solve(x, y): if maze[x][y] == 0 or maze[x][y] == 2: # 为了走动并探索迷宫 print('Visiting/Visitando xy({},{})'.format(x, y)) maze[x][y] = 2 if x < (len(maze) - 1) and (maze[x + 1][y] == 0 or maze[x + 1][y] == 3): solve(x + 1, y) elif y < (len(maze) - 1) and (maze[x][y + 1] == 0 or maze[x][y + 1] == 3): solve(x, y + 1) elif x > 0 and (maze[x - 1][y] == 0 or maze[x][y + 1] == 3): solve(x - 1, y) elif y > 0 and (maze[x][y - 1] == 0 or maze[x][y + 1] == 3): solve(x, y - 1) else: # 如果它在死胡同中卡住 step_back = 1 dead_end = True # 每种可能的死胡同类型和返回的策略 if (maze[x + 1][y] == 1 or maze[x + 1][y] == 2) and maze[x][y - 1] == 1 and \ maze[x][y + 1] == 1 and maze[x - 1][y] == 2: print('Dead end in/Caminho sem saída encontrado em xy({},{})'.format(x, y)) while dead_end is True: if maze[x - step_back][y - 1] == 0 or maze[x - step_back][y - 1] == 3: solve(x - step_back, y - 1) elif maze[x - step_back][y + 1] == 0 or maze[x - step_back][y + 1] == 3: solve(x - step_back, y + 1) else: print("Going back/Voltando") dead_end = False step_back += 1 solve(x - step_back, y) elif (maze[x - 1][y] == 1 or maze[x - 1][y] == 2) and maze[x][y - 1] == 1 and \ maze[x][y + 1] == 1 and maze[x + 1][y] == 2: print('Dead end in/Caminho sem saída encontrado em xy({},{})'.format(x, y)) while dead_end is True: if maze[x + step_back][y - 1] == 0 or maze[x - step_back][y - 1] == 3: solve(x + step_back, y - 1) elif maze[x + step_back][y + 1] == 0 or maze[x - step_back][y + 1] == 3: solve(x + step_back, y + 1) else: print("Going back/Voltando") dead_end = False step_back += 1 solve(x + step_back, y) elif (maze[x][y + 1] == 1 or maze[x][y + 1] == 2) and maze[x + 1][y] == 1 and \ maze[x - 1][y] == 1 and maze[x][y - 1] == 2: print('Dead end in/Caminho sem saída encontrado em xy({},{})'.format(x, y)) while dead_end is True: if maze[x][y - step_back] == 0 or maze[x][y - step_back] == 3: solve(x - step_back, y) elif maze[x][y + step_back] == 0 or maze[x][y + step_back] == 3: solve(x + step_back, y) else: print("Going back/Voltando") dead_end = False step_back += 1 solve(x, y + step_back) elif (maze[x][y - 1] == 1 or maze[x][y - 1] == 2) and maze[x + 1][y] == 1 and \ maze[x - 1][y] == 1 and maze[x][y + 1] == 2: print('Dead end in/Caminho sem saída encontrado em xy({},{})'.format(x, y)) while dead_end is True: if maze[x][y - step_back] == 0 or maze[x][y - step_back] == 3: solve(x - step_back, y) elif maze[x][y + step_back] == 0 or maze[x][y + step_back] == 3: solve(x + step_back, y) else: print("Going back/Voltando") dead_end = False step_back += 1 solve(x, y - step_back) # 检查是否找到了迷宫的终点 if maze[x + 1][y] == 3 or maze[x - 1][y] == 3 or maze[x][y + 1] == 3 or maze[x][y - 1] == 3: print('Exit found in/Saída encontrada em xy({},{})'.format(x, y))solve(1,1)回答：
一个简单的广度优先搜索/深度优先搜索就足以解决这个问题。从初始位置开始，并跟踪所有已覆盖的节点。如果到达任何死胡同或任何位置重复，你可以终止这条路径。如果到达最终状态，输出当前路径。
你可以在这里找到关于此算法的更多信息这里。
