我一直在研究这三个，并在下面说明了他们的推论。有人可以告诉我我是否足够正确地理解它们？谢谢。
当所有节点中的任何一个都可以作为源时，将使用
(这是最重要的一个。我的意思是，这是我最不确定的一个:)
3，贝尔曼·福特(Bellman-Ford)和迪杰斯特拉(Dijkstra)一样，只有一个来源。这可以处理负的权重，的工作方式与Floyd-Warshall的相同，除了一个来源，对吗？
如果需要看一下，相应的算法是(由维基百科提供):
贝尔曼福特:
procedure BellmanFord(list vertices, list edges, vertex source)
// This implementation takes in a graph, represented as lists of vertices
// and edges, and modifies the vertices so that their distance and
// predecessor attributes store the shortest paths.
// Step 1: initialize graph
for each vertex v in vertices:
if v is source then v.distance := 0
else v.distance := infinity
v.predecessor := null
// Step 2: relax edges repeatedly
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
// Step 3: check for negative-weight cycles
for each edge uv in edges:
u := uv.source
v := uv.destination
if u.distance + uv.weight < v.distance:
error "Graph contains a negative-weight cycle"
Dijkstra:
1 function Dijkstra(Graph, source):
2 for each vertex v in Graph: // Initializations
3 dist[v] := infinity ; // Unknown distance function from
4 // source to v
5 previous[v] := undefined ; // Previous node in optimal path
6 // from source
7
8 dist[source] := 0 ; // Distance from source to source
9 Q := the set of all nodes in Graph ; // All nodes in the graph are
10 // unoptimized - thus are in Q
11 while Q is not empty: // The main loop
12 u := vertex in Q with smallest distance in dist[] ; // Start node in first case
13 if dist[u] = infinity:
14 break ; // all remaining vertices are
15 // inaccessible from source
16
17 remove u from Q ;
18 for each neighbor v of u: // where v has not yet been
19 removed from Q.
20 alt := dist[u] + dist_between(u, v) ;
21 if alt < dist[v]: // Relax (u,v,a)
22 dist[v] := alt ;
23 previous[v] := u ;
24 decrease-key v in Q; // Reorder v in the Queue
25 return dist;
弗洛伊德·沃希尔(Floyd-Warshall):
1 /* Assume a function edgeCost(i,j) which returns the cost of the edge from i to j
2 (infinity if there is none).
3 Also assume that n is the number of vertices and edgeCost(i,i) = 0
4 */
5
6 int path[][];
7 /* A 2-dimensional matrix. At each step in the algorithm, path[i][j] is the shortest path
8 from i to j using intermediate vertices (1..k−1). Each path[i][j] is initialized to
9 edgeCost(i,j).
10 */
11
12 procedure FloydWarshall ()
13 for k := 1 to n
14 for i := 1 to n
15 for j := 1 to n
16 path[i][j] = min ( path[i][j], path[i][k]+path[k][j] );
最佳答案
您对前两个问题以及Floyd-Warshall的目标(找到所有对之间的最短路径)是正确的，但对于Bellman-Ford和Floyd-Warshall之间的关系却不正确:两种算法都使用动态编程来找到最短的路径，但FW与从每个起始节点到每个其他节点运行BF并不相同。
在BF中，问题是:从源到目标的最短路径是最多k个步骤，运行时间为O(EV)。如果我们将其运行到其他节点，则运行时间将为O(EV ^ 2)。
在FW中，问题是:对于所有节点i，j，k，从i到j到k的最短路径是什么。这导致O(V ^ 3)的运行时间-对于每个起始节点，其运行时间均优于BF(对于密集图，其运行时间高达| V |)。
关于负循环/权重的另一点注意:Dijkstra可能根本无法给出正确的结果。 BF和FW不会失败-它们会正确指出没有最小重量路径，因为负重量是无穷大的。