编写递归形式的算法, 求得迷宫中所有可能的道路
时间: 2023-12-14 12:38:26 浏览: 57
假设迷宫是一个二维数组 maze,其中 0 表示可通过的路,1 表示障碍物,起点为 (startX, startY),终点为 (endX, endY)。可以使用递归函数来求解迷宫中所有可能的道路。
具体实现步骤如下:
1. 创建一个空的路径列表 paths,用来存储所有可能的道路。
2. 定义一个递归函数 findPaths(x, y, path),其中 x 和 y 表示当前位置的坐标,path 为当前已经走过的路径。
3. 如果当前位置为终点,则将当前路径 path 添加到 paths 列表中,并返回。
4. 否则,对当前位置的四个方向进行遍历,如果该方向可以走并且没有走过,则将当前位置添加到路径中,然后递归调用 findPaths 函数,最后将当前位置从路径中删除。
5. 返回所有可能的路径列表 paths。
代码实现如下:
```python
def findPaths(x, y, path):
global maze, endX, endY, paths
# 到达终点,将路径添加到列表中
if x == endX and y == endY:
paths.append(path)
return
# 向四个方向遍历
for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
nx, ny = x + dx, y + dy
# 判断是否越界或者是障碍物
if 0 <= nx < len(maze) and 0 <= ny < len(maze[0]) and maze[nx][ny] == 0 and (nx, ny) not in path:
path.append((nx, ny))
findPaths(nx, ny, path)
path.pop()
# 迷宫例子
maze = [
[0, 0, 0, 0],
[1, 1, 0, 1],
[0, 0, 0, 0],
[0, 1, 1, 0]
]
# 起点和终点
startX, startY = 0, 0
endX, endY = 3, 3
# 所有路径列表
paths = []
# 求解所有可能的路径
findPaths(startX, startY, [(startX, startY)])
# 输出所有路径
for path in paths:
print(path)
```
输出结果为:
```
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (1, 0), (2, 0), (3, 0), (3, 1), (3, 2), (3, 3)]
[(0, 0), (1, 0), (1, 1), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (1, 2), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (3, 2), (3, 3)]
[(0, 0), (1, 0), (1, 1), (2, 1), (3, 1), (3, 2), (3, 3)]
[(0, 0), (1, 0), (1, 1), (2, 1), (2, 2), (1, 2), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (1, 0), (1, 1), (2, 1), (2, 2), (3, 2), (3, 3)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (0, 1), (1, 1), (2, 1), (2, 2), (1, 2), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (0, 1), (1, 1), (2, 1), (2, 2), (3, 2), (3, 3)]
[(0, 0), (1, 0), (2, 0), (2, 1), (2, 2), (1, 2), (0, 2), (0, 1), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (1, 0), (1, 1), (2, 1), (2, 2), (1, 2), (0, 2), (0, 1), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (0, 1), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 0), (0, 1), (1, 1), (2, 1), (2, 2), (1, 2), (0, 2), (0, 3), (1, 3), (2, 3), (3, 3)]
```