如果你改变的 Bellman-Ford算法​​一点点就可以实现的东西非常相似：

If you alter the second step of the Bellman-Ford algorithm a little bit you can achieve something very similar:

for i from 1 to size(vertices)-1:
   for each edge uv in edges: // uv is the edge from u to v
       u := uv.source
       v := uv.destination
       if u.distance + uv.weight < v.distance:
           v.distance := u.distance + uv.weight
           v.predecessor[] := u
       else if u.distance + uv.weight == v.distance:
           if u not in v.predecessor:
               v.predecessor += u

其中，诉predecessor ​​是顶点的列表。如果新的距离为V 等于它不包括尚未包括新的predecessor的路径。

where v.predecessor is a list of vertices. If the new distance of v equals a path which isn't included yet include the new predecessor.

为了打印您可以使用类似

In order to print all shortest paths you could use something like

procedure printPaths(vertex current, vertex start, list used, string path):
    if current == start:
        print start.id + " -> " + path
    else:
        for each edge ve in current.predecessors:
            if ve.start not in used:
                printPaths(ve.start,start, used + ve.start, ve.start.id + " -> " + path)

使用 printPaths（停止，启动，停止，stop.id）以打印所有路径。

注：可以排除如果u在V predecessor则v predecessor ​​+ = U 从改进算法，如果你删除。之后，该算法已经完成重复的元素。

Note: It is possible to exclude if u not in v.predecessor then v.predecessor += u from the modified algorithm if you remove duplicate elements after the algorithm has finished.