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python - 优化Python中的广度优先搜索

转载 作者:行者123 更新时间:2023-12-01 04:25:02 24 4
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我编写了一个深度优先搜索,它返回找到目标节点的深度,如果没有找到路径,则返回 -1。该算法有效,但我需要加快速度。这是函数

def depth(dic, head, target):
if(head==target):
return
depth=1
que = deque()
que.append('|') #used to mark end of each breadth so i can count depth correctly
used = list()
add = True

while(que): #while the que isn't empty
for x in dic[head]: #check current level
if x==target:
print(depth),
return;
for x in dic[head]: #add this level to the que and used list
for y in used:
if y==x:
add=False
break
if add == True:
que.append(x)
used.append(x)
add=True
que.append('|') #add our delimiter
while(que): #grab the next item from the que and inc depth count
temp = que.popleft()
if temp=='|': #bump depth counter since we found end of a level
depth+=1
else:
head=temp #bump to next node to check
break

print('-1'), #reached the end, node not found

声明传入的 dic

dic = defaultdict(list)

这样每个值都是一个整数列表,我正在使用 |所以我知道什么时候该调整深度计数器。我意识到我陷入了中间,我正在检查当前级别的所有节点并将它们添加到队列和使用列表中,但我不知道如何加快速度。

编辑:

对于任何有类似问题的人来说,这里是我最终得到的算法,它逐步执行搜索的方式有点奇怪,因为我返回的是可以找到值的最浅深度,如果不是全部的话同时检查相同深度的连接,我们最终可能会找到下一个深度的节点(就像一个错误)

def depthOfNodeBFS(dic, head, target):
if(head==target):
return
depth=0
que = []
que.append(head)
used = set()
nex = list()

while(que): #while the que isn't empty
depth+=1 #bump the depth counter
for x in que: #check the next level of all nodes at current depth
if target in dic[x]: #if the target is found were done
print(depth),
return;
else: #other wise track what we've checked
nex.append(x)
used.add(x)
while(nex): #remove checked from que and add children to que
que.pop(0)
que.extend(dic[nex.pop()]-used)
print('-1'),

最佳答案

对我来说,这看起来更像是广度优先搜索而不是深度优先搜索,但您不应该嵌套 while 循环。广度优先搜索的常用算法如下:

add the root to the queue
while the queue is not empty:
dequeue the first element elt
if elt is the solution, return it
otherwise, add each child of elt to the queue

如果你想报告深度,我建议将元组(节点,深度)添加到队列中:

add (root, 0) to the queue
while the queue is not empty:
elt, depth = queue.popleft()
if elt is the solution, return (elt, depth)
otherwise, for each child of elt:
add (child, depth+1) to the queue

关于python - 优化Python中的广度优先搜索,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33237105/

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