用连续的字母字符标记迷宫解决方案路径

Question

我正在研究迷宫解算器，虽然我的所有功能都可以正常工作，但我无法让一部分工作。我需要用字母表 (a-z) 的升序字母而不是单个字符来标记 'solution path'。

我无法让它继续遍历字母字符列表，我取得的最好成绩是它停在字母 'a' 处并将所有字母更改为该字母。

迷宫：

###_###
#_____#
#_##_##
#_##__#
#_#####

'solved maze':

###a###
#aaa__#
#a##_##
#a##__#
#a#####

代码：

PATH = "_"
START = "_"
VISITED = "."
SOLUTION = "o"

class Maze:
    def __init__(self, ascii_maze):
        # splits maze string into separate values on each new line
        # uses list comp. to create 'matrix' out of maze
        self.maze = [list(row) for row in ascii_maze.splitlines()]
        # finds index first '_' character which denotes the beginning
        self.start_y = [row.count(START) for row in self.maze].index(1)
        # finds position where '_' is located within the 'y' line which returns the index position
        self.start_x = self.maze[self.start_y].index(START)

    # returns string representation of maze object, joining maze elements passed in as parameters
    # used to print maze with 'cells' and be user friendly
    def __repr__(self):
        return "\n".join("".join(row) for row in self.maze)

    def solve_maze(self, x=None, y=None):
        # assigns starting (x,y) position
        if x is None:
            x, y = self.start_x, self.start_y
        # checks if the coordinate is in the path/start
        if self.maze[y][x] in (PATH, START):
            # marks spot as 'visited' for recursion check
            self.maze[y][x] = VISITED
            # uses recursion to check paths by checking each direction and making a decision off that
            try:
                if (self.solve_maze(x+1, y) or
                        self.solve_maze(x-1, y) or
                        self.solve_maze(x, y+1) or
                        self.solve_maze(x, y-1)):
                    self.maze[y][x] = SOLUTION
                    return True
            # this exception is what occurs when the program tries to leave the maze (aka it found the exit)
            # also marks it
            except IndexError:
                self.maze[y][x] = SOLUTION
                return True
        return False


if __name__ == "__main__":
    import sys
    alphabet_list = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k','l', 'm', 'n',
                     'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
    # for checking mazes through terminal
    if len(sys.argv) > 1:
        maze = Maze(open(sys.argv[1]).read())
    # if none available, defaults to 'maze' text file
    else:
        maze = Maze(open('maze').read())
    # prints string representation of maze to replace "visited" areas (. char used for testing) with original "_"
    if maze.solve_maze():
        # converting to string allows for easy replacement of things
        maze = str(maze)
        maze = maze.replace(".", "_")
        for char in maze:
            if char == "o":
                for item in alphabet_list:
                    maze = maze.replace(char, item)

    print(maze)

Answer 1

由于您需要将解决方案字符设置为已解决（事后通过不知情的字符串操作很难重新创建它，因此它基本上是 re-navigating 整个解决方案路径）：

我们将使用辅助函数 return 为我们提供字母表中的下一个字符，重置为 'z':

SOLUTION = ord('a') - 1
def getSolutionChar():
    global SOLUTION
    if SOLUTION >= ord('z'):
        SOLUTION = ord('a')
    else:
        SOLUTION += 1
    return chr(SOLUTION)

try 块的替换：

        try:
            if (self.solve_maze(x+1, y) or
                self.solve_maze(x-1, y) or
                self.solve_maze(x, y+1) or
                self.solve_maze(x, y-1)):
                self.maze[y][x] = getSolutionChar()
                return True
        except IndexError:
            self.maze[y][x] = getSolutionChar()
            return True

不要忘记删除以下内容，因为不再需要它：

alphabet_list = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k','l', 'm', 'n',
                 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

    for char in maze:
        if char == "o":
            for item in alphabet_list:
                maze = maze.replace(char, item)

用连续的字母字符标记迷宫解决方案路径

Mark maze solution path with consecutive alphabet characters

python

for-loop

maze