Algorithm 宽度优先算法的实现

我试图实现一个“广度优先”算法，作为我在一本书中看到的东西的一个变体

我的问题是算法没有将每个节点的元素添加到队列中

例如，如果我在“search（）”函数中搜索名为“mariela”的“black lab”，我将得到正确的输出：“simon是一个black lab”

然而，我应该能够在“沃尔特”中找到“黑色实验室”，它与“玛丽拉”相连，玛丽拉与“西蒙”相连，西蒙是一个“黑色实验室”。这是行不通的

我是否在实现这个算法时犯了新手错误，或者我的图形设置错误

一如既往，我们非常感谢您的任何帮助

from collections import deque


# TEST GRAPH -------------
graph = {}
graph['walter'] = ['luci', 'kaiser', 'andrea', 'mariela']
graph['andrea'] = ['echo', 'dante', 'walter', 'mariela']
graph['mariela'] = ['ginger', 'simon', 'walter', 'andrea']
graph['kaiser'] = 'german shepherd'
graph['luci'] = 'black cat'
graph['echo'] = 'pitbull'
graph['dante'] = 'pitbull'
graph['ginger'] = 'orange cat'
graph['simon'] = 'black lab'



def condition_met(name):
    if graph[name] == 'black lab':
        return name


def search(name):
    search_queue = deque()
    search_queue += graph[name]             # add all elements of    "name" to queue
    searchedAlready = []                    # holding array for people    already searched through

while search_queue:                     # while queue not empty...
    person = search_queue.popleft()     # pull 1st person from queue
    if person not in searchedAlready:   # if person hasn't been searched through yet...
        if condition_met(person):
            print person + ' is a black labrador'
            return True
    else:
        search_queue += graph[person]
        searchedAlready.append(person)
return False


 search('walter')
#search('mariela')

---

在您的实现中有很多问题，包括Python和算法方面的问题

重写为：

# @param graph  graph to search
# @param start  the node to start at
# @param value  the value to search for
def search(graph, start, value):
  explored = []
  queue = [start]
  while len(queue) > 0:
    # next node to explore
    node = queue.pop()
    # only explore if not already explored
    if node not in explored:
      # node found, search complete
      if node == value:
        return True
      # add children of node to queue
      else:
        explored.append(node)
        queue.extend(graph[node]) # extend is faster than concat (+=)
  return False


graph = {}
graph['walter'] = ['luci', 'kaiser', 'andrea', 'mariela']
graph['andrea'] = ['echo', 'dante', 'walter', 'mariela']
graph['mariela'] = ['ginger', 'simon', 'walter', 'andrea']

# children should be a list
graph['kaiser'] = ['german shepherd'] 
graph['luci'] = ['black cat']
graph['echo'] = ['pitbull']
graph['dante'] = ['pitbull']
graph['ginger'] = ['orange cat']
graph['simon'] = ['black lab']

print search(graph, 'mariela', 'walter')

---

这是一个演示

谢谢。我来试一试。我已经用--print search（图形“沃尔特”、“德国牧羊犬”）运行了你的代码--它没有返回正确的输出。我收到了一个错误。原因是没有名为“pitbull”的节点，即使您有名为

pitbull

的子节点，您的图形中也没有名为“pitbull”的节点a）

图形['pitbull']=[]

或B）更改此行

队列。扩展（图形[节点]）

to

queue.extend（graph.get（node，[]））