使用堆和队列数据结构解决迷宫问题

 

 

python实现迷宫问题的栈和队列的解决方法：

#迷宫问题
#表示迷宫的颜色，0表示路通，1表示围墙
maze=[
    [1,1,1,1,1,1,1,1,1,1],
    [1,0,0,1,0,0,0,1,0,1],
    [1,0,0,1,0,0,0,1,0,1],
    [1,0,0,0,0,1,1,0,0,1],
    [1,0,1,1,1,0,0,0,0,1],
    [1,0,0,0,1,0,0,0,0,1],
    [1,0,1,0,0,0,1,0,0,1],
    [1,0,1,1,1,0,1,1,0,1],
    [1,1,0,0,0,0,0,0,0,1],
    [1,1,1,1,1,1,1,1,1,1]
]

d=[
    lambda x,y:(x+1,y),
    lambda x,y:(x-1,y),
    lambda x,y:(x,y+1),
    lambda x,y:(x,y-1)
    ]

#方法1：使用栈来进行解决，方法是回溯法，即深度优先搜索，但是并非是最佳最短的路线
def maze_path(x1,y1,x2,y2):
    stack=[]
    stack.append((x1,y1))
    while(len(stack)>0):
        curnode=stack[-1]
        if curnode[0]==x2 and curnode[1]==y2:
            for path in stack:
                print(path)
            return True
        for i in d:
            nextnode=i(curnode[0],curnode[1])
            #如果下一个位置可以走
            if maze[nextnode[0]][nextnode[1]]==0:
                stack.append(nextnode)
                maze[nextnode[0]][nextnode[1]]= 2  #2表示走过了
                break
        else:
            maze[nextnode[0]][nextnode[1]] =2
            stack.pop()
    else:
        print("没路")
        return False
maze_path(1,1,8,8)

maze=[
    [1,1,1,1,1,1,1,1,1,1],
    [1,0,0,1,0,0,0,1,0,1],
    [1,0,0,1,0,0,0,1,0,1],
    [1,0,0,0,0,1,1,0,0,1],
    [1,0,1,1,1,0,0,0,0,1],
    [1,0,0,0,1,0,0,0,0,1],
    [1,0,1,0,0,0,1,0,0,1],
    [1,0,1,1,1,0,1,1,0,1],
    [1,1,0,0,0,0,0,0,0,1],
    [1,1,1,1,1,1,1,1,1,1]
]
#解决方法2：使用队列来进行记录，属于广度优先搜索
from collections import deque  #导入队列的内置模块
#输出路径函数
def print_r(path):
    real_path=[]
    i=len(path)-1
    while i>=0:
        real_path.append(path[i][0:2])
        i=path[i][2]
    real_path.reverse()
    for p in real_path:
        print(p)

def maze_path1(x1,y1,x2,y2):
    quede=deque()
    path=[]
    quede.append((x1,y1,-1))
    while len(quede)>0:
        curnode=quede.popleft()
        path.append(curnode)
        if curnode[0]==x2 and curnode[1]==y2:
            #到达终点
            print_r(path)
        for dir in d:
            nextnode=dir(curnode[0],curnode[1])
            if maze[nextnode[0]][nextnode[1]]==0:
                quede.append((nextnode[0],nextnode[1],len(path)-1))
                maze[nextnode[0]][nextnode[1]] =2 #标记已经走过
    return False

maze_path1(1,1,8,8)