Python 最短距离算法

我想创建一个简单的广度优先搜索算法，它返回最短路径

for i in movie_info:
    actor_info[i] = movie_info[i]

def find_shortest_path(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return path
    if not start in graph:
        return None
    shortest = None
    for node in graph[start]:
        if node not in path:
            newpath = find_shortest_path(graph, node, end, path)
            if newpath:
                if not shortest or len(newpath) < len(shortest):
                    shortest = newpath
    return shortest

L = find_shortest_path(actor_info, 'act1', 'act2')
print len([i for i in L if i in movie_info])

演员信息字典将演员和演员映射到演员出现的电影列表：

actor_info = { "act1" : ["movieC", "movieA"], "act2" : ["movieA", "movieB"], 
     "act3" :["movieA", "movieB"], "act4" : ["movieC", "movieD"], 
     "act5" : ["movieD", "movieB"], "act6" : ["movieE"], 
     "act7" : ["movieG", "movieE"], "act8" : ["movieD", "movieF"], 
     "KevinBacon" : ["movieF"], "act10" : ["movieG"], "act11" : ["movieG"] }

与此相反，电影将映射到其中出现的演员列表：

movie_info = {'movieB': ['act2', 'act3', 'act5'], 'movieC': ['act1', 'act4'], 
      'movieA': ['act1', 'act2', 'act3'], 'movieF': ['KevinBacon', 'act8'], 
      'movieG': ['act7', 'act10', 'act11'], 'movieD': ['act8', 'act4', 'act5'], 
      'movieE': ['act6', 'act7']}

打个电话

shortest_dictance("act1", "Kevin Bacon", actor_info, movie_info)

我应该得到

3

，因为

act1

出现在

movieC

和

Act4

一起出现在

movieD

和

Act8

一起出现在

movief

和

KevinBacon

一起。所以最短的距离是3

到目前为止，我有：

def shotest_distance(actA, actB, actor_info, movie_info):
   '''Return the number of movies required to connect actA and actB. 
   If theres no connection return -1.'''

    # So we keep 2 lists of actors:
    #   1.The actors that we have already investigated.
    #   2.The actors that need to be investigated because we have found a 
    #      connection beginning at actA. This list must be 
    #      ordered, since we want to investigate actors in the order we 
    #      discover them.
    #      -- Each time we put an actor in this list, we also store
    #         her distance from actA.
    investigated = []
    to_investigate = [actA]
    distance = 0
    while actB not in to_investigate and to_investigate!= []:
        for actor in to_investigate:
            to_investigated.remove(actA)
            investigated.append(act)

            for movie in actor_info[actor]:
                for co_star in movie_info[movie]:
                    if co_star not in (investigated and to_investigate):
                        to_investigate.append(co_star)

 ....
 ....

 return d    

---

我无法找到适当的方法来跟踪在每次代码迭代中发现的距离。此外，代码在时间上似乎非常无效。

首先从中创建一个图形来连接所有节点，然后运行最短路径代码（可能有一个高效的图形库来执行此操作，而不是下面提到的函数，不过这是一个优雅的图形库）然后从最短路径中找出所有电影名称的数目

for i in movie_info:
    actor_info[i] = movie_info[i]

def find_shortest_path(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return path
    if not start in graph:
        return None
    shortest = None
    for node in graph[start]:
        if node not in path:
            newpath = find_shortest_path(graph, node, end, path)
            if newpath:
                if not shortest or len(newpath) < len(shortest):
                    shortest = newpath
    return shortest

L = find_shortest_path(actor_info, 'act1', 'act2')
print len([i for i in L if i in movie_info])

电影中的i的

信息：
演员信息[i]=电影信息[i]
def查找最短路径（图形、开始、结束、路径=[]）：
路径=路径+[开始]
如果开始=结束：
返回路径
如果未在图形中启动：
一无所获
最短=无
对于图形[start]中的节点：
如果节点不在路径中：
newpath=查找最短路径（图形、节点、端点、路径）
如果是新路径：
如果不是最短的或len（新路径）

---

查找最短路径来源：

这看起来很有效。它跟踪当前的一组电影。对于每一步，它都会查看所有尚未被考虑（“看过”）的一步之遥电影

输出是

All movies with act1 set(['movieC', 'movieA'])
Movies with actors from those movies set(['movieB', 'movieC', 'movieA', 'movieD'])
New movies with actors from those movies set(['movieB', 'movieD'])
Movies with actors from those movies set(['movieB', 'movieC', 'movieA', 'movieF', 'movieD'])
New movies with actors from those movies set(['movieF'])
3

这是一个版本，它返回构成最小连接的电影列表（无连接为无，如果actA和actB相同，则为空列表）

以下是输出：

The Kevin Bacon Number for act1 is 3
Movies are: movieC, movieD, movieF

---

编码nit：在图形中使用

start

而不是

graph.has_key（start）

。这是面试问题吗？

The Kevin Bacon Number for act1 is 3
Movies are: movieC, movieD, movieF