如何在广度优先搜索中跟踪路径? [英] How to trace the path in a Breadth-First Search?

问题描述

如何跟踪广度优先搜索的路径，例如在以下示例中:

How do you trace the path of a Breadth-First Search, such that in the following example:

如果搜索键11，则返回连接1到11的最短列表.

If searching for key 11, return the shortest list connecting 1 to 11.

[1, 4, 7, 11]

推荐答案

你应该看看 http://en.wikipedia.org/wiki/Breadth-first_search 首先.

下面是一个快速实现，其中我使用了一个列表列表来表示路径队列.

Below is a quick implementation, in which I used a list of list to represent the queue of paths.

# graph is in adjacent list representation
graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

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, '1', '11')

这会打印:['1', '4', '7', '11']

另一种方法是维护从每个节点到其父节点的映射，并在检查相邻节点时记录其父节点.搜索完成后，只需根据父映射回溯即可.

Another approach would be maintaining a mapping from each node to its parent, and when inspecting the adjacent node, record its parent. When the search is done, simply backtrace according the parent mapping.

graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def backtrace(parent, start, end):
    path = [end]
    while path[-1] != start:
        path.append(parent[path[-1]])
    path.reverse()
    return path
        

def bfs(graph, start, end):
    parent = {}
    queue = []
    queue.append(start)
    while queue:
        node = queue.pop(0)
        if node == end:
            return backtrace(parent, start, end)
        for adjacent in graph.get(node, []):
            if node not in queue :
                parent[adjacent] = node # <<<<< record its parent 
                queue.append(adjacent)

print bfs(graph, '1', '11')

以上代码基于没有循环的假设.

The above codes are based on the assumption that there's no cycles.

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