使用BFS查找从起点到终点的所有路径

问题描述

使用下面的BFS实现，我们如何修改程序以存储从开始到结束节点的所有路径？有什么想法吗？

def bfs(graph_to_search,start,end):
    queue = [[start]]
    visited = set()

    while queue:
        # Gets the first path in the queue
        path = queue.pop(0)

        # Gets the last node in the path
        vertex = path[-1]

        # Checks if we got to the end
        if vertex == end:
            return path
        # We check if the current node is already in the visited nodes set in order not to recheck it
        elif vertex not in visited:
            # enumerate all adjacent nodes,construct a new path and push it into the queue
            for current_neighbour in graph_to_search.get(vertex,[]):
                new_path = list(path)
                new_path.append(current_neighbour)
                queue.append(new_path)

            # Mark the vertex as visited
            visited.add(vertex)

解决方法

也许类似的事情可能起作用。我认为，在检查是否访问了某个节点时，您会从结果中排除许多路径，这仅仅是因为它们包含相同的节点。我将其更改为将整个路径添加到已访问，然后改为选中此路径。同样，创建一个空列表results，如果找到最后一个节点，我们会附加该列表。您的代码将在找到最后一个节点的那一刻从函数中返回，并且不会浏览并找到剩余的路径。让我知道这是否有效！

def bfs(graph_to_search,start,end):
    queue = [[start]]
    visited = set()
    results = []

    while queue:
        # Gets the first path in the queue
        path = queue.pop(0)

        # Gets the last node in the path
        vertex = path[-1]

        # Checks if we got to the end
        if vertex == end:
            results.append(path)
            continue
        # We check if the current path is already in the visited nodes set in order not to recheck it
        elif path not in visited:
            # enumerate all adjacent nodes,construct a new path and push it into the queue
            for current_neighbour in graph_to_search.get(vertex,[]):
                new_path = path.copy()
                new_path.append(current_neighbour)
                queue.append(new_path)

            # Mark the vertex as visited
            visited.add(path)
      return results

您还可以将queue中的元素另存为包含当前顶点和先前路径的元组，如下所示。可能更具可读性。

queue = [(start,[])]
vertex,path = queue.pop(0)