题解 | #迷宫问题#

inline1 = input().split(' ')
rows, cols = int(inline1[0]), int(inline1[1])
maze = []
for _ in range(rows):
    inline2 = list(map(int, input().split(' ')))
    maze.append(inline2)

from typing import Any


# rows, cols = 5, 5
# maze = [[0, 1, 0, 0, 0],
#         [0, 1, 1, 1, 0],
#         [0, 0, 0, 0, 0],
#         [0, 1, 1, 1, 0],
#         [0, 0, 0, 1, 0],]

class Solution:
    def __init__(self, maze_in, rows_in, cols_in) -> None:
        self.maze = maze_in
        self.rows = rows_in
        self.cols = cols_in
        self.path = []
    
    def explore(self, start):
        if start == [self.rows-1, self.cols-1]:
            return True
        
        # 以下每一种搜索路径不冲突，不能用elif
        # 如果有抵达路径，则一定会在以下判断中返回True，反之会走完以下所有判断返回False
        if 0 < start[0] and not self.maze[start[0]-1][start[1]]:
            self.maze[start[0]-1][start[1]] = 2
            if self.explore([start[0]-1, start[1]]):
                self.path.append([start[0]-1, start[1]])
                return True
        if start[0] < self.rows-1 and not self.maze[start[0]+1][start[1]]:
            self.maze[start[0]+1][start[1]] = 2
            if self.explore([start[0]+1, start[1]]):
                self.path.append([start[0]+1, start[1]])
                return True
        if 0 < start[1] and not self.maze[start[0]][start[1]-1]:
            self.maze[start[0]][start[1]-1] = 2
            if self.explore([start[0], start[1]-1]):
                self.path.append([start[0], start[1]-1])
                return True
        if start[1] < self.cols-1 and not self.maze[start[0]][start[1]+1]:
            self.maze[start[0]][start[1]+1] = 2
            if self.explore([start[0], start[1]+1]):
                self.path.append([start[0], start[1]+1])
                return True
        return False

    def __call__(self, start) -> Any:
        if self.maze[start[0]][start[1]] == 1:  # 判断0,0是否有可行路径
            return False
        else:
            self.maze[start[0]][start[1]] = 2
            if self.explore(start):
                self.path.append(start)
                for ptr in self.path[::-1]:
                    print('({},{})'.format(*ptr))
            else:
                return False


sol = Solution(maze, rows, cols)
sol([0, 0])

关键在于递归终止条件的写法，任一个路径如果递归抵达返回True，所有路径都不能递归抵达则返回False