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Python利用Prim算法生成迷宫

作者:Leleprogrammer

普里姆算法(Prim算法),图论中的一种算法,可在加权连通图里搜索最小生成树。这篇文章将利用Prim算法实现迷宫的生成,感兴趣的可以了解一下

Prim算法随机生成后的迷宫数列矩阵如下图:

15x15:

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 s 0 1 0 1 0 1 1 1 1 1 0 1 0 
0 1 0 1 0 1 0 0 0 1 0 0 0 1 0 
0 1 1 1 1 1 1 1 1 1 1 1 0 1 0 
0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 
0 1 0 1 1 1 0 1 1 1 0 1 0 1 0 
0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 
0 1 1 1 1 1 1 1 1 1 1 1 1 1 0 
0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 
0 1 0 1 0 1 1 1 0 1 1 1 1 1 0 
0 1 0 1 0 0 0 1 0 0 0 1 0 1 0 
0 1 0 1 0 1 1 1 1 1 0 1 0 1 0 
0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 
0 1 1 1 0 1 1 1 1 1 1 e 0 1 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

27x27:

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 
0 s 1 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 1 1 1 1 0 
0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 
0 1 1 1 0 1 0 1 1 1 0 1 0 1 1 1 1 1 0 1 0 1 1 1 1 1 0 
0 0 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 
0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 0 1 0 1 1 1 0 
0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 1 0 0 0 
0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 1 0 1 0 1 0 1 0 1 1 1 0 
0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 
0 1 0 1 1 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 
0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 
0 1 0 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 0 1 0 
0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 
0 1 1 1 1 1 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 1 0 1 0 1 0 
0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 1 0 
0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 
0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 1 0 1 0 
0 1 1 1 1 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 1 0 1 0 
0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 0 0 0 0 0 
0 1 1 1 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 1 0 1 1 1 1 1 0 
0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 1 0 1 0 
0 1 1 1 1 1 1 1 0 1 0 1 1 1 0 1 1 1 1 1 0 1 0 1 0 1 0 
0 1 0 1 0 1 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 
0 1 0 1 0 1 0 1 0 1 1 1 1 1 0 1 1 1 1 1 0 1 0 1 1 1 0 
0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 0 0 0 0 
0 1 1 1 e 1 0 1 0 1 1 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 0 
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 

实现该效果的代码如下: 

import random as rd
 
nearmaybe=[
    [-2,0],
    [2,0],
    [0,-2],
    [0,2]
]
 
def createMaze(row,col):
    maze=[[0 for i in range(col)] for i in range(row)]
    check=[]
    firstrow=rd.randrange(1,row-2,2)
    firstcol=rd.randrange(1,col-2,2)
    maze[firstrow][firstcol]=1
    check.append([firstrow,firstcol])
    while len(check):
        c=rd.choice(check)
        nears=[]
        conditions=[]
        for maybe in nearmaybe:
            conditions.append([c[0]+maybe[0],c[1]+maybe[1]])
        for condition in conditions:
            if condition[0]>=1 and condition[0]<=row-2 \
                    and condition[1]>=1 and condition[1]<=col-2:
                nears.append([condition[0],condition[1]])
        for n in nears.copy():
            if maze[n[0]][n[1]]:
                nears.remove(n)
        for block in nears:
            if block[0]==c[0]:
                if block[1]<c[1]:
                    maze[block[0]][c[1]-1]=1
                    maze[block[0]][block[1]]=1
                    check.append([block[0],block[1]])
                else:
                    maze[block[0]][block[1]-1]=1
                    maze[block[0]][block[1]]=1
                    check.append([block[0],block[1]])
            else:
                if block[0]<c[0]:
                    maze[c[0]-1][block[1]]=1
                    maze[block[0]][block[1]]=1
                    check.append([block[0],block[1]])
                else:
                    maze[block[0]-1][block[1]]=1
                    maze[block[0]][block[1]]=1
                    check.append([block[0],block[1]])
        if not len(nears):
            check.remove(c)
    maze[1][1]="s"
    while True:
        c=rd.randint(1,col-2)
        if maze[row-2][c]:
            maze[row-2][c]="e"
            break
 
    return maze

调用该函数的方法:

maze=createMaze(27,27)
for l in maze:
    for s in l:
        print(s,end=" ")
    print()

CreateMaze传入的两个参数必须都是单数,否则程序可能会无法运行!

下面的for循环遍历就是将生成的迷宫矩阵打印出来

S表示出发点

1表示路

0表示墙

E表示终点

生成迷宫矩阵后,就可以将其引用到游戏中去啦!

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