如何使用BFS按顺序获取包含某些给定节点的路径？

Question

我有一个带有未加权边缘的图表，其中每个节点都标有字母'a'到'z'。

我想修改BFS算法以获得包含字母'c'，'o'，'d'，'e'的最短路径。这四个人之间可能还有其他字母。您有起始节点'a'和结束节点'b'。您可以假设始终是包含该顺序中的四个字母的路径。如何修改BFS以满足该条件？

Answer 1

如果您知道如何使用BFS在两个节点之间找到最短路径，那么问题可以解决如下：

• 找到从 a 到 c
的最短路径
• 查找从 c 到 o 的最短路径
• 找出从 o 到 d
的最短路径
• 查找从 d 到 e
的最短路径
• 找出从 e 到 b
的最短路径
• 连接上述5条路径，从而产生从 a 到 b 的一条路径。

这是Python中的一个实现：

class Node:
    def __init__(self, name):
        self.name = name
        self.neighbors = []

    def link(self, node): 
        # The edge is undirected: implement it as two directed edges
        self.neighbors.append(node)
        node.neighbors.append(self)

    def shortestPathTo(self, target):
        # A BFS implementation which retains the paths
        queue = [[self]]
        visited = set()
        while len(queue):
            path = queue.pop(0) # Get next path from queue (FIFO)
            node = path[-1] # Get last node in that path
            for neighbor in node.neighbors:
                if neighbor == target:
                    # Found the target node. Return the path to it
                    return path + [target]
                # Avoid visiting a node that was already visited
                if not neighbor in visited:
                    visited.add(neighbor)
                    queue.append(path + [neighbor])

# Create the nodes of the graph (indexed by their names)
graph = {}
for letter in 'abcdefo':
    graph[letter] = Node(letter)

# Create the undirected edges
for start, end in ['ab','ae','bc','be','bo','df','ef','fo']:
    graph[start].link(graph[end])

# Concatenate the shortest paths between each of the required node pairs 
start = 'a'
path = [graph['a']]
for end in ['c','o','d','e','b']:
    path.extend( graph[start].shortestPathTo(graph[end])[1:] )
    start = end

# Print result: the names of the nodes on the path
print([node.name for node in path])

代码中创建的图形如下所示：

输出结果为：

['a', 'b', 'c', 'b', 'o', 'f', 'd', 'f', 'e', 'b']