如何使用Python NetworkX找到最长的路径?

时间:2014-08-31 06:53:05

标签: python networkx

我有从S到T的有向图。

我想找到路线(S,A,C,E,T)及其容量之和(1 + 2 + 3 + 1 = 7),因此总和最大。

我尝试过networkx.algorithms.flow.ford_fulkerson,但我不知道如何从S到T获得单向指示。

我的环境:

  • Ubuntu 12.04
  • Python 2.7.8
  • NetworkX 1.9
  • Matplotlib 1.4.0

example.py

#!/usr/bin/env python
# -*- coding: utf-8 -*-

import matplotlib.pylab as p
import networkx as nx

if __name__ == '__main__':
    DG = nx.DiGraph()
    DG.add_edge('S', 'a', capacity=1)
    DG.add_edge('a', 'b', capacity=1)
    DG.add_edge('a', 'c', capacity=2)
    DG.add_edge('b', 'd', capacity=1)
    DG.add_edge('b', 'e', capacity=2)
    DG.add_edge('c', 'e', capacity=3)
    DG.add_edge('c', 'f', capacity=2)
    DG.add_edge('d', 'T', capacity=1)
    DG.add_edge('e', 'T', capacity=1)
    DG.add_edge('f', 'T', capacity=1)

    result = nx.algorithms.flow.ford_fulkerson(DG, 'S', 'T')
    print(result.size(weight='capacity')) # 15.0, but I want 7.

    pos = nx.spectral_layout(DG)
    nx.draw(DG, pos)
    nx.draw_networkx_labels(DG, pos)
    nx.draw_networkx_edge_labels(DG, pos)
    p.show()

    # This shows a partly bidirectional graph, which is not what I want.
    pos = nx.spectral_layout(result)
    nx.draw(result, pos)
    nx.draw_networkx_labels(result, pos)
    nx.draw_networkx_edge_labels(result, pos)
    p.show()

3 个答案:

答案 0 :(得分:1)

使用负权重通常不适用于Dijkstra算法。 此错误ValueError: ('Contradictory paths found:', 'negative weights?')将会显示。

它应该区分“最长路径”和“最大总和路径”的问题。

答案在这里:How to find path with highest sum in a weighted networkx graph?,使用all_simple_paths

请注意,在函数all_simple_paths(G, source, target, cutoff=None)中,使用cutoff param(整数)可以帮助限制从sourcetarget的搜索深度。它还控制我们想要找到的路径的长度。

答案 1 :(得分:1)

负重量适用于约翰逊。在您的情况下,修改为:

DG = nx.DiGraph()
DG.add_edge('S', 'a', weight=-1)
DG.add_edge('a', 'b', weight=-1)
DG.add_edge('a', 'c', weight=-2)
DG.add_edge('b', 'd', weight=-1)
DG.add_edge('b', 'e', weight=-2)
DG.add_edge('c', 'e', weight=-3)
DG.add_edge('c', 'f', weight=-2)
DG.add_edge('d', 'T', weight=-1)
DG.add_edge('e', 'T', weight=-1)
DG.add_edge('f', 'T', weight=-1)

要找到最长的路径,请使用

获取从S到T的单向指示
p2 = nx.johnson (DG, weight='weight')
print('johnson: {0}'.format(p2['S']['T']))

结果为:johnson: ['S', 'a', 'c', 'e', 'T']

我的环境:

  • 软件版本
  • Python 3.4.5 64bit [MSC v.1600 64 bit(AMD64)]
  • IPython 5.1.0 OS Windows 10 10.0.14393
  • networkx 1.11

Networkx 1.11 docs for johnson

答案 2 :(得分:0)

我想我找到了解决方案。

#!/usr/bin/env python
# -*- coding: utf-8 -*-

import networkx as nx

def inverse_weight(graph, weight='weight'):
    copy_graph = graph.copy()
    for n, eds in copy_graph.adjacency_iter():
        for ed, eattr in eds.items():
            copy_graph[n][ed][weight] = eattr[weight] * -1
    return copy_graph

def longest_path_and_length(graph, s, t, weight='weight'):
    i_w_graph = inverse_weight(graph, weight)
    path = nx.dijkstra_path(i_w_graph, s, t)
    length = nx.dijkstra_path_length(i_w_graph, s, t) * -1
    return path, length

if __name__ == '__main__':
    DG = nx.DiGraph()
    DG.add_edge('S', 'a', weight=1)
    DG.add_edge('a', 'b', weight=1)
    DG.add_edge('a', 'c', weight=2)
    DG.add_edge('b', 'd', weight=1)
    DG.add_edge('b', 'e', weight=2)
    DG.add_edge('c', 'e', weight=3)
    DG.add_edge('c', 'f', weight=2)
    DG.add_edge('d', 'T', weight=1)
    DG.add_edge('e', 'T', weight=1)
    DG.add_edge('f', 'T', weight=1)

    print(longest_path_and_length(DG, 'S', 'T')) # (['S', 'a', 'c', 'e', 'T'], 7)