我一直在研究这三个问题,下面我将从中得出推论。有人能告诉我我对它们的理解是否足够准确吗?非常感谢。
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Dijkstra的算法只有当你有一个单一的源,并且你想知道从一个节点到另一个节点的最小路径时才使用,但在这种的情况下失败了
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当所有节点中的任何一个都可以是源时,都会使用Floyd-Warshall的算法,因此您希望从任何源节点到达任何目的节点的最短距离。只有当出现负循环时,此操作才会失败
(这是最重要的一个。我的意思是,这是我最不确定的一个:)
3.当只有一个来源时,Bellman-Ford的用法就像Dijkstra的一样。这可以处理负权重,除了一个来源外,它的工作原理与Floyd Warshall的相同,对吧
如果你需要看看,相应的算法是(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:
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] );
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Bellman Ford和Floyd-Warrshall之间的关系不是正确的:两种算法都使用动态编程来找到最短路径,但FW与从每个起始节点到其他节点运行BF不同。
在BF中,问题是:最多使用k步,从源到目标的最短路径是什么,运行时间是O(EV)。如果我们将它运行到每个其他节点,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通过k的最短路径是什么。这导致O(V^3)的运行时间——对于每个起始节点都比BF好(对于密集图,高达|V|的因子)。
关于负循环/权重还有一点需要注意:Dijkstra可能根本无法给出正确的结果。BF和FW不会失败——它们会正确地声明没有最小权重路径,因为负权重是无限的。
单源最短路径:
Dijkstra算法-不允许负权重-O(E+Vlg(V))
Bellman-ford算法-允许负权重。但如果存在负循环,Bellman-ford将检测到-ve循环-O(ve)
有向无循环图-顾名思义,它只适用于DAG-O(V+E)
所有对的最短路径:
Dijkstra算法-不允许负权重-O(VE+V^2lg(V))
Bellman-ford算法-O(V^2E)
矩阵链乘法方法-复杂度与Bellman-ford算法相同
Floyd-Warshall算法-使用动态编程方法-复杂性为O(V^3)