乘以Networkx圖邊的算法

Question

所以我的問題是在用Networkx庫實現的圖中找到從一個節點到另一個節點（或同一節點）的最長路徑。

我不想增加邊緣的權重，但要乘以最大的結果。 顯然，每個節點僅傳遞一次或根本不傳遞。

例如，如果我要從節點1轉到節點4，則最佳結果將是：2 x 14 x 34 x 58

圖示例

謝謝您的幫助 ！

Answer 1

這可能起作用：

import networkx as nx

G = nx.Graph()

# create the graph
G.add_edge(1, 2, weight=2 )
G.add_edge(1, 4, weight=5 )
G.add_edge(2, 3, weight=14 )
G.add_edge(2, 4, weight=5 )
G.add_edge(2, 5, weight=4 )
G.add_edge(3, 5, weight=34 )
G.add_edge(4, 5, weight=58 )

start = 1 # start node
end = 4   # end node

all_paths = [path for path in nx.all_simple_paths(G, start, end)]

# initialize result
largest_path = None
largest_path_weight = None

# iterate through all paths to find the largest
for p in all_paths:                                       # keep track of each path
    for _ in range(len(p)):                               # for each node in this path
        pairs = zip(p, p[1:])                             # get sequence of nodes
        product = 1                                       # reset product for this paths calculation
        for pair in pairs:                                # for each pair of nodes in this path
            an_edge = G.get_edge_data(pair[0], pair[1])   # get this edge's data
            product *= an_edge['weight']                  # multiply all weights
    if product > largest_path_weight:                     # check if this path's product is greater
        largest_path = p                                  # if True, set largest to this path
        largest_path_weight = product                     # save the weight of this path

# display result
print 'largest path:', largest_path 
print 'weight:', largest_path_weight

對於此示例：

largest path: [1, 2, 3, 5, 4]
weight: 55216