我已经学习了这三个算法，并在下面陈述了我的推论。请问是否有人能告诉我我是否理解得足够准确？谢谢。
Dijkstra算法仅在您有一个单一源并且想要知道从一个节点到另一个节点的最短路径时使用，但在像这样的情况下会失败。
Floyd-Warshall算法用于任何节点都可以是源头的情况，因此您希望找到从任何源节点到达任何目标节点的最短距离。只有当存在负循环时，该算法才会失败。
(这是最重要的一个。我的意思是，我对这个问题最不确定:)
3.Bellman-Ford算法与Dijkstra类似，用于仅有一个源头的情况。它可以处理负权重，除了只有一个源头之外，其工作方式与Floyd-Warshall相同，对吗？
如果您需要查看相应的算法，请参考（由维基百科提供）：
Bellman-Ford：
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;
Floyd-Warshall算法：
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] );