關(guān)于迷宮問題，常見會問能不能到達(dá)某點(diǎn)，以及打印到達(dá)的最短路徑。可以用回溯法+遞歸來解決

代碼一：dfs + 回溯

將矩陣外圍以及路障統(tǒng)一設(shè)置為-1

設(shè)置current_step和next_step兩個值。如果面臨的點(diǎn)大于next_step，或者從未走過，就可以對它遞歸

存儲矩陣：直接在原迷宮上修改，每個點(diǎn)標(biāo)記的是從起點(diǎn)過來經(jīng)過的步數(shù)，遇到0或者是比當(dāng)前大至少2的，就考察

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

def init():

global graph

graph.append([-1, -1, -1, -1, -1, -1, -1])

graph.append([-1, 0, 0, -1, 0, 0, -1])

graph.append([-1, 0, -1, 0, 0, 0, -1])

graph.append([-1, 0, -1, 0, -1, -1, -1])

graph.append([-1, 0, -1, 0, 0, 0, -1])

graph.append([-1, 0, 0, 0, -1, 0, -1])

graph.append([-1, 0, 0, 0, 0, 0, -1])

graph.append([-1, -1, -1, -1, -1, -1, -1])

#深度優(yōu)先遍歷

def deepFirstSearch( steps , x, y ):

global graph

current_step = steps + 1

print(x, y, current_step )

graph[x][y] = current_step

next_step = current_step + 1

'''

遍歷周圍4個點(diǎn):

如果周圍節(jié)點(diǎn)不是-1 說明 不是障礙 在此基礎(chǔ)上：

里面是0 說明沒遍歷過 我們把它修改成當(dāng)前所在位置步數(shù)加1

里面比當(dāng)前的next_step大 說明不是最優(yōu)方案 就修改它

里面比當(dāng)前next_step說明當(dāng)前不是最優(yōu)方案，不修改

'''

if not(x-1== 1 and y==1) and graph[x-1][y] != -1 and ( graph[x-1][y]>next_step or graph[x-1][y] ==0 ) : #上

deepFirstSearch(current_step, x-1 , y )

if not(x == 1 and y-1==1) and graph[x][y-1] != -1 and ( graph[x][y-1]>next_step or graph[x][y-1] ==0 ) : #左

deepFirstSearch(current_step, x , y-1 )

if not(x == 1 and y+1==1) and graph[x][y+1] != -1 and ( graph[x][y+1]>next_step or graph[x][y+1]==0 ) : #右

deepFirstSearch(current_step, x , y+1 )

if not(x+1== 1 and y==1) and graph[x+1][y] != -1 and ( graph[x+1][y]>next_step or graph[x+1][y]==0 ) : #下

deepFirstSearch(current_step, x+1 , y )

if __name__ == "__main__":

graph = []

init()

deepFirstSearch(-1,1,1)

print(graph[1][5])

代碼二：也是回溯

這里0是可走，1是路障，用while來做循環(huán)。如果某點(diǎn)的附近有判斷可走的新點(diǎn)，就將新點(diǎn)append進(jìn)stack，走過了標(biāo)記為-1；如果四個方向都沒有，就從舊點(diǎn)從stack中pop掉，標(biāo)上-1，進(jìn)行回溯。

存儲矩陣：stack中存儲的0是可走，1是路障，-1是走過的或者是死路

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],

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

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

]

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 mpath(x1, y1, x2, y2):

stack = []

stack.append((x1, y1))

while len(stack) > 0:

curNode = stack[-1]

if curNode[0] == x2 and curNode[1] == y2:

#到達(dá)終點(diǎn)

for p in stack:

print(p)

return True

for dir in dirs:

nextNode = dir(curNode[0], curNode[1])

if maze[nextNode[0]][nextNode[1]] == 0:

#找到了下一個

stack.append(nextNode)

maze[nextNode[0]][nextNode[1]] = -1 # 標(biāo)記為已經(jīng)走過，防止死循環(huán)

break

else:#四個方向都沒找到

maze[curNode[0]][curNode[1]] = -1 # 死路一條,下次別走了

stack.pop() #回溯

print("沒有路")

return False

mpath(1,1,8,8)

總結(jié)

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