networkx中具有“关键”节点可访问的最短路径

Question

我在用networkx开发的python G中有一个有向图。 该图的权重称为“权重”。

我知道显式的起始节点A和结束节点F。在图之间可以访问节点B，C，D，E。

我该如何明确地说他必须通过找到最短路径来访问B和D，并且可以附加地添加C和E（如果这有助于最短路径）？

到目前为止，我知道函数：

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

它给出了输出：

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

我如何确保它包含E？

Answer 1

Networkx没有针对您的问题的内置函数或参数。 您应该手动进行：

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)))