Question

问题是找到迷宫的出口。我在这段代码中找不到错误。迷宫编码：1 - 坑（或墙，它取决于），0 - 高速公路，2,3 - 访问的字段。 S - 南（下），E - 东（左），N - 北（上），W - 西（左）。 lepes()是执行递归移动的主要函数。 x - 活动移动的水平，y - 垂直坐标。顺便说一下，迷宫的长度是12 x 12，被坑（墙壁，默认这意味着1）所包围。真正的游乐场是一个10 x 10的场地。所有访问的移动存储在列表s中，最初为s = []。变量lab存储迷宫。我只是试图找出迷宫的出路。我在Python中有这个代码：

def lepes(x, y, lab, s):
    if x != 10 and y !=10:
        # step forward... 
        lab[x][y] = 3
        # can I move down?
        if x < 11 and lab[x+1][y] == 0 :
            s.append("S")
            lepes(x+1, y, lab, s)
        # can I move right?
        if y < 11 and lab[x][y+1] == 0:
            s.append("E")
            lepes(x, y+1, lab, s)
        # can I move up?
        if x > 0 and lab[x-1][y] == 0:
            s.append("N")
            lepes(x-1, y, lab, s)        
        # can I move left?
        if y > 0 and lab[x][y-1] == 0:
            s.append("W")
            lepes(x, y-1, lab, s)
        #   step back...
        #   mark as visited
        #lab[x][y] = 2
        s.append("")
        #s.pop()
    else:
        # The goal is reached, and last step forward...
        lab[x][y] = 3
        return
        # last step back 
        lab[x][y] = 2

为了找到迷宫的出路，我试图从初始点（1,1）调用函数lepes(1, 1, lab, s)。我必须用坐标（10,10）到达现场：使用此初始值lab：

lab = [[1,1,1,1,1,1,1,1,1,1,1,1],[1,0,0,0,0,0,0,0,0,0,0,1],[1,0,1,1,1,1,1,1,0,1,1,1],[1,0,1,0,0,0,0,0,0,0,0,1],[1,0,1,0,1,1,1,1,1,1,0,1],[1,0,1,0,1,0,0,0,0,0,0,1],[1,0,0,0,1,1,0,1,1,1,0,1],[1,0,1,0,0,0,0,1,0,1,1,1],[1,0,1,1,0,1,0,0,0,0,0,1],[1,0,1,0,0,1,1,1,1,1,0,1],[1,0,0,0,1,1,0,0,0,0,0,1],[1,1,1,1,1,1,1,1,1,1,1,1]]

最终解决方案形式："".join(s) 我有这个：

s =“SSSSSSSSSEESSESSWSEESEEEENNNEEEEWNNNEEEEEEENNEEWWWWWW”

我应该有这样的事情：

s =“SSSSSEENNNEEEEEEESSWWWWSSSEEEESS”

Maze to solve

黄色是起点，绿色是目标。

Answer 1

如果您正在寻找最短路径，我建议如下：

将您的迷宫转换为具有以下属性的加权图：

顶点集是所有可通过的正方形的集合。
边缘集是相邻可通行方块的所有元组的集合。
每条边的重量为1。

在此之后让Dijkstra先生或A *为你做这项工作。

我能找到的最短路径是“SSSSSEESEEESEEEESS”。

这里是我发现它的快速和脏代码：

#! /usr/bin/python3

lab = [[1,1,1,1,1,1,1,1,1,1,1,1],[1,0,0,0,0,0,0,0,0,0,0,1],[1,0,1,1,1,1,1,1,0,1,1,1],[1,0,1,0,0,0,0,0,0,0,0,1],[1,0,1,0,1,1,1,1,1,1,0,1],[1,0,1,0,1,0,0,0,0,0,0,1],[1,0,0,0,1,1,0,1,1,1,0,1],[1,0,1,0,0,0,0,1,0,1,1,1],[1,0,1,1,0,1,0,0,0,0,0,1],[1,0,1,0,0,1,1,1,1,1,0,1],[1,0,0,0,1,1,0,0,0,0,0,1],[1,1,1,1,1,1,1,1,1,1,1,1]]

class Node:
    def __init__ (self, x, y):
        self.x = x
        self.y = y
        self.neighbours = [ (x + xoff, y + yoff) for xoff, yoff in
            ( (1, 0), (0, 1), (0, -1), (-1, 0) )
            if not lab [y + yoff] [x + xoff] ]
        self.distance = ...
        self.path = ...
        self.visited = False

    def __repr__ (self):
        return '{}: ({})'.format ( (self.x, self.y), self.neighbours)

nodes = {}
for y in range (12):
    for x in range (12):
        if lab [y] [x]: continue
        nodes [x, y] = Node (x, y)

current = nodes [1, 1]
current.distance = 0
current.path = []
unvisited = set (nodes.keys () )

while True:
    dist = current.distance + 1
    for nx, ny in current.neighbours:
        if (nx, ny) not in unvisited: continue
        neighbour = nodes [nx, ny]
        if neighbour.distance is ... or neighbour.distance > dist:
            neighbour.distance = dist
            neighbour.path = current.path + [ (current.x, current.y) ]
    current.visited = True
    unvisited.remove ( (current.x, current.y) )
    if not unvisited: break
    current = sorted ( [node for node in nodes.values ()
        if not node.visited and node.distance is not ...],
        key = lambda node: node.distance) [0]

print (nodes [10, 10].path)
path = nodes [10, 10].path + [ (10, 10) ]
for (ax, ay), (bx, by) in zip (path, path [1:] ):
    if ax == bx and ay > by: print ('N', end = '')
    if ax == bx and ay < by: print ('S', end = '')
    if ay == by and ax > bx: print ('W', end = '')
    if ay == by and ax < bx: print ('E', end = '')
print ()

结果是：

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 6), (3, 6), (3, 7), (4, 7), (5, 7), (6, 7), (6, 8), (7, 8), (8, 8), (9, 8), (10, 8), (10, 9)]
SSSSSEESEEESEEEESS

或者如果您在右上角开始，结果是：

[(10, 1), (9, 1), (8, 1), (8, 2), (8, 3), (9, 3), (10, 3), (10, 4), (10, 5), (9, 5), (8, 5), (7, 5), (6, 5), (6, 6), (6, 7), (6, 8), (7, 8), (8, 8), (9, 8), (10, 8), (10, 9)]
WWSSEESSWWWWSSSEEEESS

Answer 2

您的if语句不是唯一的。在每个函数内部，您可以递归地调用函数，但是当递归调用返回时，执行只会继续执行，并且可以进入同一位置的其他块之一。

您可以通过将它们更改为elif来解决此问题，但我个人并不认为递归是最佳解决方案（除非您明确尝试以函数式编程）：最好在顶部使用while循环，并在if的每个分支内更新x和y。

在Python中找到迷宫中的出路

2 个答案: