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我有一个脚本，它使用邻接图和 BFS 算法来查找两点之间的路径。该图有大约 10,000 个顶点，脚本设置如下：

graph = {...
         '9660': ['9661', '9662', '9659'],
         '9661': ['9654', '9660'],
         '9662': ['9660', '9663'],
         '9663': ['9664', '9662'],
         '9664': ['9665', '9663'],
                              ...}


def bfs(graph, start, end):
# maintain a queue of paths
queue = []
# push the first path into the queue
queue.append([start])
while queue:
    # get the first path from the queue
    path = queue.pop(0)
    # get the last node from the path
    node = path[-1]
    # path found
    if node == end:
        return path
    # enumerate all adjacent nodes, construct a new path and push it into the queue
    for adjacent in graph.get(node, []):
        new_path = list(path)
        new_path.append(adjacent)
        queue.append(new_path)

print bfs(graph, '50', '8659')

因为这个算法适用于小的邻接图，我猜 python 只需要很长时间来处理这个大小的图。我的目标是找到最短的路径，但如果我什至找不到一条路径，那目前是不可能的。是否有使用大型邻接图处理路径查找的 python 解决方案？如果是这样，我可以举个例子吗？

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1 回答 1

1

您没有跟踪访问过的节点，如果您的图不是有向无环图，这可能会导致大量时间浪费。例如，如果您的图表是

{'A': ['B', 'C', 'D', 'E'],
 'B': ['A', 'C', 'D'],
 'C': ['A', 'B', 'D'],
 'D': ['A', 'B', 'C'],
 'E': ['F'],
 'F': ['G'],
 'G': ['H'],
 ...
 'W': ['X'],
 'X': ['Y'],
 'Y': ['Z']}

使用您的算法调用bfs(graph, 'A', 'Z')会在最终到达 Z 之前不必要地循环通过“A”、“B”、“C”和“D”。而如果您跟踪访问过的节点，则只需添加“A”、“B”的邻居, 'C' 和 'D' 到队列中各一次。

def bfs(graph, start, end):
    # maintain a queue of paths
    queue = []

    # push the first path into the queue
    queue.append([start])

    # already visited nodes
    visited = set()

    while queue:
        # get the first path from the queue
        path = queue.pop(0)

        # get the last node from the path
        node = path[-1]

        # if node has already been visited
        if node in visited:
            continue

        # path found
        if node == end:
            return path
        # enumerate all adjacent nodes, construct a new path and push it into the queue
        else:
            for adjacent in graph.get(node, []):
                # add the path only if it's end node hasn't already been visited
                if adjacent not in visited
                    new_path = list(path)
                    new_path.append(adjacent)
                    queue.append(new_path)

            # add node to visited set
            visited.add(node)

使用这个版本的算法和字母图，这是队列和访问集在整个算法的 while 循环顶部的样子：

queue = [ ['A'] ]
visited = {}

queue = [ ['A', 'B'], ['A', 'C'], ['A', 'D'], ['A', 'E'] ]
visited = {'A'}

queue = [ ['A', 'C'], ['A', 'D'], ['A', 'E'], ['A', 'B', 'C'],
          ['A', 'B', 'D'] ]
visited = {'A', 'B'}

queue = [ ['A', 'D'], ['A', 'E'], ['A', 'B', 'C'], ['A', 'B', 'D'],
          ['A', 'C', 'D'] ]
visited = {'A', 'B', 'C'}

queue = [ ['A', 'E'], ['A', 'B', 'C'], ['A', 'B', 'D'], ['A', 'C', 'D'] ]
visited = {'A', 'B', 'C', 'D'}

queue = [ ['A', 'B', 'C'], ['A', 'B', 'D'], ['A', 'C', 'D'], ['A', 'E', 'F'] ]
visited = {'A', 'B', 'C', 'D', 'E'}

queue = [ ['A', 'B', 'D'], ['A', 'C', 'D'], ['A', 'E', 'F'] ]
visited = {'A', 'B', 'C', 'D', 'E'}

queue = [ ['A', 'C', 'D'], ['A', 'E', 'F'] ]
visited = {'A', 'B', 'C', 'D', 'E'}

queue = [ ['A', 'E', 'F'] ]
visited = {'A', 'B', 'C', 'D', 'E'}

queue = [ ['A', 'E', 'F', 'G'] ]
visited = {'A', 'B', 'C', 'D', 'E', 'F'}

queue = [ ['A', 'E', 'F', 'G', 'H'] ]
visited = {'A', 'B', 'C', 'D', 'E', 'F', 'G'}

...
...

queue = [ ['A', 'E', 'F', 'G', 'H', ..., 'X', 'Y', 'Z'] ]
visited = {'A', 'B', 'C', 'D', 'E', 'F', 'G', ..., 'X', 'Y'}

# at this point the algorithm will pop off the path,
# see that it reaches the goal, and return

这比添加路径要少得多['A', 'B', 'A', 'B', 'A', 'B', ...]。

于 2012-12-04T12:27:35.147 回答