网络中的最短路径，要访问'key'节点

Question

我有一个用networkx开发的python G有向图。该图具有称为"权重"的权重。

我知道一个明确的开始节点 A 和一个结束节点 F。在图之间可以访问节点B，C，D，E。

我怎么能明确地说他必须通过找到最短路径来访问 B 和 D，并且可以加法添加 C 和 E，如果这有助于最短路径？

到目前为止，我知道该功能：

nx.single_source_dijkstra(G, 'A', target='F', cutoff=None, weight='weight')

给出输出：

(10.01,
['A',
 'B',
 'C',
 'F',])

如何确保它包含 E？

Answer 1

Networkx没有针对您的问题的内置函数或参数。您应该手动执行此操作：

import networkx as nx
# Create a random DAG
G = nx.gnp_random_graph(50,0.3,directed=True)
DAG = nx.DiGraph([(u,v) for (u,v) in G.edges() if u<v])
nx.is_directed_acyclic_graph(DAG)
for edge in G.edges:
    G.edges[edge]['weight'] = 1
# Get the longest path (without weights) from node 1 to node 40
# with nodes 5, 10, 20, 30 inside
max([
    (path, len(path))
    for path in nx.all_simple_paths(DAG, 1, 40)
    if all(n in path for n in (5, 10, 20, 30))
], key=lambda x: x[1])
# Get the longest path (with weights)
max([
    path
    for path in nx.all_simple_paths(DAG, 1, 40)
    if all(n in path for n in (5, 10, 20, 30))
], key=lambda x: sum(G.edges[edge]['weight'] for edge in nx.utils.pairwise(x)))