python - Igraph/networkx 中的 k 最短路径实现(Yen 算法)

Question

经过深入研究并基于this , this还有很多建议我实现 k 最短路径算法，以便在大型无向循环加权图中找到第一、第二、第三……第 k 条最短路径。大约2000个节点。

Wikipedia 上的伪代码这是:

function YenKSP(Graph, source, sink, K):
  //Determine the shortest path from the source to the sink.
 A[0] = Dijkstra(Graph, source, sink);
 // Initialize the heap to store the potential kth shortest path.
 B = [];

for k from 1 to K:
   // The spur node ranges from the first node to the next to last node in the shortest path.
   for i from 0 to size(A[i]) − 1:

       // Spur node is retrieved from the previous k-shortest path, k − 1.
       spurNode = A[k-1].node(i);
       // The sequence of nodes from the source to the spur node of the previous k-shortest path.
       rootPath = A[k-1].nodes(0, i);

       for each path p in A:
           if rootPath == p.nodes(0, i):
               // Remove the links that are part of the previous shortest paths which share the same root path.
               remove p.edge(i, i) from Graph;

       // Calculate the spur path from the spur node to the sink.
       spurPath = Dijkstra(Graph, spurNode, sink);

       // Entire path is made up of the root path and spur path.
       totalPath = rootPath + spurPath;
       // Add the potential k-shortest path to the heap.
       B.append(totalPath);

       // Add back the edges that were removed from the graph.
       restore edges to Graph;

   // Sort the potential k-shortest paths by cost.
   B.sort();
   // Add the lowest cost path becomes the k-shortest path.
   A[k] = B[0];
return A;

主要问题是我还不能为此编写正确的 python 脚本(删除边并将它们正确地放回原位)所以我像往常一样依赖 Igraph 只能走到这一步:

def yenksp(graph,source,sink, k):
    global distance
    """Determine the shortest path from the source to the sink."""
    a = graph.get_shortest_paths(source, sink, weights=distance, mode=ALL, output="vpath")[0]
    b = [] #Initialize the heap to store the potential kth shortest path
    #for xk in range(1,k):
    for xk in range(1,k+1):
        #for i in range(0,len(a)-1):
        for i in range(0,len(a)):
            if i != len(a[:-1])-1:
                spurnode = a[i]
                rootpath = a[0:i]
                #I should remove edges part of the previous shortest paths, but...:
                for p in a:
                    if rootpath == p:
                        graph.delete_edges(i) 

            spurpath = graph.get_shortest_paths(spurnode, sink, weights=distance, mode=ALL, output="vpath")[0]
            totalpath = rootpath + spurpath
            b.append(totalpath)
            # should restore the edges
            # graph.add_edges([(0,i)]) <- this is definitely not correct.
            graph.add_edges(i)
        b.sort()
        a[k] = b[0]
    return a

这是一个非常糟糕的尝试，它只返回一个列表中的一个列表

我不太确定我在做什么，我已经对这个问题感到非常绝望，在过去的几天里，我对此的看法发生了 180 度的转变，甚至一次。 我只是一个尽力而为的菜鸟。请帮忙。也可以建议 Networkx 实现。

附言很可能没有其他可行的方法来解决这个问题，因为我们已经在这里进行了研究。我已经收到了很多建议，我欠社区很多。 DFS 或 BFS 不会工作。图很大。

编辑:我一直在更正 python 脚本。简而言之，这个问题的目的是正确的脚本。

Answer 1

Github 上有 Yen 的 KSP 的 python 实现，YenKSP .充分感谢作者，此处给出了算法的核心:

def ksp_yen(graph, node_start, node_end, max_k=2):
    distances, previous = dijkstra(graph, node_start)

    A = [{'cost': distances[node_end], 
          'path': path(previous, node_start, node_end)}]
    B = []

    if not A[0]['path']: return A

    for k in range(1, max_k):
        for i in range(0, len(A[-1]['path']) - 1):
            node_spur = A[-1]['path'][i]
            path_root = A[-1]['path'][:i+1]

            edges_removed = []
            for path_k in A:
                curr_path = path_k['path']
                if len(curr_path) > i and path_root == curr_path[:i+1]:
                    cost = graph.remove_edge(curr_path[i], curr_path[i+1])
                    if cost == -1:
                        continue
                    edges_removed.append([curr_path[i], curr_path[i+1], cost])

            path_spur = dijkstra(graph, node_spur, node_end)

            if path_spur['path']:
                path_total = path_root[:-1] + path_spur['path']
                dist_total = distances[node_spur] + path_spur['cost']
                potential_k = {'cost': dist_total, 'path': path_total}

                if not (potential_k in B):
                    B.append(potential_k)

            for edge in edges_removed:
                graph.add_edge(edge[0], edge[1], edge[2])

        if len(B):
            B = sorted(B, key=itemgetter('cost'))
            A.append(B[0])
            B.pop(0)
        else:
            break

    return A