算法之迷宫问题
题目：给一个二维列表，表示迷宫（0表示通道，1表示围墙）。给出算法，求一条走出迷宫的路径。
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] ]
解决思路：
- 在一个迷宫节点(x,y)上，可以进行四个方向的探查：maze[x-1][y], maze[x+1][y], maze[x][y-1], maze[x][y+1]
- 思路：从一个节点开始，任意找下一个能走的点，当找不到能走的点时，退回上一个点寻找是否有其他方向的点。
- 方法：创建一个空栈，首先将入口位置进栈。当栈不空时循环：获取栈顶元素，寻找下一个可走的相邻方块，如果找不到可走的相邻方块，说明当前位置是死胡同，进行回溯（就是讲当前位置出栈，看前面的点是否还有别的出路）
用栈实现：
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] ] dirs = [ lambda x,y:(x-1,y), #上 lambda x,y:(x,y+1), #右 lambda x,y:(x+1,y), #下 lambda x,y:(x,y-1), #左 ] def solve_maze(x1, y1, x2, y2): stack = [] stack.append((x1,y1)) maze[x1][y1] = 2 while len(stack) > 0: # 当栈不空循环 cur_node = stack[-1] if cur_node == (x2,y2): #到达终点 for p in stack: print(p) return True for dir in dirs: next_node = dir(*cur_node) if maze[next_node[0]][next_node[1]] == 0: #找到一个能走的方向 stack.append(next_node) maze[next_node[0]][next_node[1]] = 2 # 2表示已经走过的点 break else: #如果一个方向也找不到 stack.pop() else: print("无路可走") return False
用队列实现：
解决思路：
- 思路：从一个节点开始，寻找所有下面能继续走的点。继续寻找，直到找到出口。
- 方法：创建一个空队列，将起点位置进队。在队列不为空时循环：出队一次。如果当前位置为出口，则结束算法；否则找出当前方块的4个相邻方块中可走的方块，全部进队。
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] ] def solve_maze2(x1,y1,x2,y2): queue = deque() path = [] # 记录出队之后的节点 queue.append((x1,y1,-1)) maze[x1][y1] = 2 while len(queue) > 0: cur_node = queue.popleft() path.append(cur_node) if cur_node[0] == x2 and cur_node[1] == y2: #到终点 real_path = [] x,y,i = path[-1] real_path.append((x,y)) while i >= 0: node = path[i] real_path.append(node[0:2]) i = node[2] real_path.reverse() for p in real_path: print(p) return True for dir in dirs: next_node = dir(cur_node[0], cur_node[1]) if maze[next_node[0]][next_node[1]] == 0: queue.append((next_node[0], next_node[1], len(path)-1)) maze[next_node[0]][next_node[1]] = 2 # 标记为已经走过 else: print("无路可走") return False solve_maze2(1,1,8,8)
