通向终点的路不止一条！python迷宫。

wei_Y

其实木有上面那么高端啦。。只是数字版。。

规则:

1. 迷宫没有墙壁，但是每条边的路被坑包围。如果一个玩家掉在坑里的话，他们就丢失了。迷宫是用矩阵来表示（含有列表的列表）: 1是坑，0的路径的一部分。 迷宫的大小是12×12，外界都是坑。玩家在点（1,1）开始。出口是在点（10,10）。 你需要找到通过迷宫的路径。 玩家只能通过四个方向移动 南（下 [1,0]），北（上 [-1,0]）， 东（右[0,1]），西（左[0，-1]）。这条路线被描述成由不同的字符组成的字符串：“S”=南，“N”=北，“E”=东，和“W”=西。

复制代码

迷宫:
One:

[

[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],

]

Two:

[

[1,1,1,1,1,1,1,1,1,1,1,1],

[1,0,1,0,0,0,1,0,0,0,0,1],

[1,0,1,0,1,0,1,0,1,1,0,1],

[1,0,1,0,1,0,1,0,1,0,0,1],

[1,0,1,0,1,0,1,0,1,0,1,1],

[1,0,1,0,1,0,1,0,1,0,0,1],

[1,0,1,0,1,0,1,0,1,1,0,1],

[1,0,1,0,1,0,1,0,1,0,0,1],

[1,0,1,0,1,0,1,0,1,0,1,1],

[1,0,1,0,1,0,1,0,1,0,0,1],

[1,0,0,0,1,0,0,0,1,1,0,1],

[1,1,1,1,1,1,1,1,1,1,1,1],

]

three:

[

[1,1,1,1,1,1,1,1,1,1,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,0,1],

[1,1,1,1,1,1,1,1,1,1,1,1],

]

Four:

[

[1,1,1,1,1,1,1,1,1,1,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,0,1],

[1,1,1,1,1,1,1,1,1,1,1,1],

]

Five:

[

[1,1,1,1,1,1,1,1,1,1,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,1,1],

[1,0,0,0,1,0,0,0,1,0,0,1],

[1,0,1,0,0,0,1,0,0,0,0,1],

[1,1,1,1,1,1,1,1,1,1,1,1],

]

测试代码:

1. def check_route(func, labyrinth):
   
2.         MOVE = {"S": (1, 0), "N": (-1, 0), "W": (0, -1), "E": (0, 1)}
   
3.         #copy maze
   
4.         route = func([row[:] for row in labyrinth])
   
5.         pos = (1, 1)
   
6.         goal = (10, 10)
   
7.         for i, d in enumerate(route):
   
8.             move = MOVE.get(d, None)
   
9.             if not move:
   
10.                 print("Wrong symbol in route")
    
11.                 return False
    
12.             pos = pos[0] + move[0], pos[1] + move[1]
    
13.             if pos == goal:
    
14.                 return True
    
15.             if labyrinth[pos[0]][pos[1]] == 1:
    
16.                 print("Player in the pit")
    
17.                 return False
    
18.         print("Player did not reach exit")
    
19.         return False

复制代码

带上写的函数和迷宫列表:

嘻嘻，能不能写粗来呢，各位试试吧。

我写的太菜了不贴上来丢人了。

大神写的:

游客，如果您要查看本帖隐藏内容请回复

戴宇轩

懂了!!!!! 是这样的吧?
#迷宫算法
def checkio(data): #主函数
    result=[] #结果存放处
    dirs=[[0,1,'D'],[1,0,'R'],[0,-1,'U'],[-1,0,'L']] #方向定义
    def move(path,x,y,field): #移动函数
        field[y][x]=1 #把自己变成1, 防止无限递归
        if x==10 and y==10: #如果到终点了
            result.append(path) #将路径放入结果
        for d in dirs: #检查四个方向
            if field[y+d[1]][x+d[0]]==0: #如果某个方向为0
                move(path+d[2],x+d[0],y+d[1],field) #递归
    move('',1,1,data) #调用移动函数, 此时路径为空
    return result[0] #将结果路径返回, Done!

来自:鱼C论坛 iPhone客户端来自: iPhone客户端

DY2

看看

Jaydonkeor

最近碰到递归一直搞不出来，拜读一下大神作品

andy大宝

没看懂

ayuanyuan

学习学习

1547042803

厉害

全体成员

哇

骑乌龟的蜗牛

练习一下

qpwoeiruyt

谢谢  学习了

夏夜星空

谢谢楼主

竟无语凝噎

为什么我在我的Python编译器里面编译完 没有任何显示

Looker

学习下

fishgang

果然大神！

MaoYi62498

前来学习一下，谢谢楼主

jerryxjr1220

果然是递归解法，思路差不多

jerryxjr1220

我感觉可以用递归解法

人zai旅途

先留下标记，后期学习

Minic

好強大的一個題目喔
我要努力學習讓自己可以寫出來這樣的編程

白鹭

初学者表示一脸蒙蔽

xfs2016

谢谢分享