使用队列解决迷宫问题(广度优先搜索 / 最短路径)

博客地址:https://www.cnblogs.com/zylyehuo/

# -*- coding: utf-8 -*-

from collections import deque

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]
]

# 上下左右四个节点
dirs = [
    lambda x, y: (x + 1, y),
    lambda x, y: (x - 1, y),
    lambda x, y: (x, y - 1),
    lambda x, y: (x, y + 1)
]


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 node in real_path:
        print(node)


def maze_path_queue(x1, y1, x2, y2):
    queue = deque()
    queue.append((x1, y1, -1))
    path = []  # 存储路径
    while len(queue) > 0:  # 当队列不空时循环
        curNode = queue.popleft()
        path.append(curNode)
        if curNode[0] == x2 and curNode[1] == y2:
            # 终点
            print_r(path)
            return True
        for dir in dirs:
            nextNode = dir(curNode[0], curNode[1])
            if maze[nextNode[0]][nextNode[1]] == 0:
                queue.append((nextNode[0], nextNode[1], len(path) - 1))  # 后续节点进队,记录哪个节点带他来的
                maze[nextNode[0]][nextNode[1]] = 2  # 标记为已经走过
    else:
        print("没有路")
        return False


maze_path_queue(1, 1, 8, 8)  # 起点坐标和终点坐标

posted @ 2023-08-17 14:29  zylyehuo  阅读(11)  评论(0编辑  收藏  举报