python - 我需要找到源节点和目标节点之间的最短路径(距离)。鉴于必须包含某些节点

Question

我需要从蓝色圆圈到达红色圆圈。该路径必须包含黑色圆圈(即使它可能不是最佳的)。

我已经包括了节点到节点的距离。通过使用“dijkstra_path”我得到:

这是正确的。

但是...我该怎么做才能确保包含“kountoumas”，甚至包含其他节点的列表。

然后运行算法或其他算法。

谢谢

Answer 1

根据我对您上一条评论的理解，您希望列出经过中间节点的所有可能路径，以便能够选择最短的路径。因此，对于图中的图表，这里是列出从 1 到 2 的所有可能路径分别作为第一个和最后一个节点，中间节点为 3 和 4 的代码。我添加了一些注释，试图使其尽可能清晰。

start = 1 # starting node for the path (fixed)
end = 2 # last node in the path (fixed)
intermediate = [4,3] # Intermediate nodes that the path must include
#permutations of intermediate nodes to find shortest path
p =  list(it.permutations(intermediate))
print "Possible orders of intermediate nodes", p, '\n'
hops_tmp = 0
path_tmp = [] # stores path for each permutation
sub_path_tmp = [] # stores sub path from one node to another
for j in xrange(len(p)): # loop for all permutations possibilities
    # path from starting node to the first intermediate node
    sub_path_tmp = nx.dijkstra_path(G,start,p[j][0]) 
    for k in xrange(len(sub_path_tmp)): # update path with sub_path
        path_tmp.append(sub_path_tmp[k])
    #loop to find path from intermediate to another upto the last node
    for i in xrange(len(intermediate)):
        # if last intermediate node calculate path to last node
        if i == len(intermediate) - 1:
            sub_path_tmp =  nx.dijkstra_path(G,p[j][i],end)
        else: # otherwise calculate path to the next intermediate node
            sub_path_tmp =  nx.dijkstra_path(G,p[j][i],p[j][i+1]) 
        for k in xrange(len(sub_path_tmp)-1): # update path with sub_path
            path_tmp.append(sub_path_tmp[k+1])
    hops_tmp = len(path_tmp) -1
    print path_tmp
    print hops_tmp , '\n'
    # Reset path and hops for the next permutation
    hops_tmp = 0
    path_tmp = []

结果如下:

 Possible orders of intermediate nodes [(4, 3), (3, 4)] 

 [1, 8, 4, 8, 1, 3, 7, 5, 9, 2]

 9

 [1, 3, 1, 8, 4, 5, 9, 2]

 7 

P.S 1- you can add other intermediate nodes if you wish and it should work

2- extracting the shortest path should be easy but I did not include it just to focus on the core of the problem