Python 利用堆栈实现深度受限路径查找
嘿,就像标题所说的,我试图在Python3中实现一个深度有限的搜索,该搜索返回给定图形的路径、起始顶点和目标顶点。我正在为如何强制执行搜索限制而挣扎。到目前为止,我已经:Python 利用堆栈实现深度受限路径查找,python,search,depth-first-search,Python,Search,Depth First Search,嘿,就像标题所说的,我试图在Python3中实现一个深度有限的搜索,该搜索返回给定图形的路径、起始顶点和目标顶点。我正在为如何强制执行搜索限制而挣扎。到目前为止,我已经: def dfs(g, v, goal, limit=-1): SENTINEL = object() visitedStack = [v] path = "" while visitedStack: currentVertex = visitedStack.pop()
def dfs(g, v, goal, limit=-1):
SENTINEL = object()
visitedStack = [v]
path = ""
while visitedStack:
currentVertex = visitedStack.pop()
if g.getVertex(currentVertex) != None:
if g.getVertex(currentVertex).visited == False:
path += currentVertex + ' -> '
g.getVertex(currentVertex).hasBeenVisited()
if currentVertex == goal:
return path[:-3]
elif currentVertex == SENTINEL:
limit += 1
elif limit != 0:
limit -= 1
visitedStack.append(SENTINEL)
visitedStack.extend(g.getVertex(currentVertex).getConnections())
return "Depth limit was reached"
编辑:我更改了一些代码以检查访问的顶点。在我编辑后,返回的搜索有时无法正常工作。例如,我将深度限制设置为3,但返回的路径为4或5。其他时候,我会将限制设置为7,并返回“已达到限制”。注意:最小路径为3当搜索深入时,将哨兵推到堆栈上并降低限制。当您从堆栈中弹出哨兵时,将增加级别
def dfs_limit(g, start, goal, limit=-1):
'''
Perform depth first search of graph g.
if limit >= 0, that is the maximum depth of the search.
'''
SENTINEL = object()
visitedStack = [start]
path = []
while visitedStack:
currentVertex = visitedStack.pop()
if currentVertex == goal:
path.append(currentVertex)
return ' -> '.join(path)
elif currentVertex == SENTINEL:
#finished this level; go back up one level
limit += 1
path.pop()
elif limit != 0:
# go one level deeper, push sentinel
limit -= 1
path.append(currentVertex)
visitedStack.append(SENTINEL)
visitedStack.extend(g.getVertex(currentVertex).getConnections())
如果图中有循环或多个路由,您还需要跟踪访问了哪些节点,这样您就不会重复工作或陷入无休止的循环。非常感谢!不过,我遇到了一些问题,我在帖子中做了一些编辑。你可以将所有未访问的节点添加到路径中,但从不删除节点。当搜索深入一层时,当前节点应添加到路径的末尾,但当搜索返回到一层时,当前节点将被删除。我修改了我的解决方案以跟踪路径。