Dijkstra's algorithm: Difference between revisions

Content added Content deleted
(→‎{{header|C++}}: simplified by using adjacency list and int to represent nodes)
(→‎{{header|OCaml}}: added translation from c++)
Line 957: Line 957:


<pre>a -> c -> d -> e</pre>
<pre>a -> c -> d -> e</pre>

{{trans|C++}}
Translation of the C++ solution, and all the complexities are the same as in the C++ solution. In particular, we again use a self-balancing binary search tree (<code>Set</code>) to implement the priority queue, which results in an optimal complexity.
<lang ocaml>type vertex = int
type weight = float
type neighbor = vertex * weight
module VertexSet = Set.Make(struct type t = weight * vertex let compare = compare end)

let dijkstra (src:vertex) (adj_list:neighbor list array) : weight array * vertex array =
let n = Array.length adj_list in
let min_distance = Array.make n infinity in
min_distance.(src) <- 0.;
let previous = Array.make n (-1) in
let rec aux vertex_queue =
if not (VertexSet.is_empty vertex_queue) then
let dist, u = VertexSet.min_elt vertex_queue in
let vertex_queue' = VertexSet.remove (dist, u) vertex_queue in
let edges = adj_list.(u) in
let f vertex_queue (v, weight) =
let dist_thru_u = dist +. weight in
if dist_thru_u >= min_distance.(v) then
vertex_queue
else begin
let vertex_queue' = VertexSet.remove (min_distance.(v), v) vertex_queue in
min_distance.(v) <- dist_thru_u;
previous.(v) <- u;
VertexSet.add (min_distance.(v), v) vertex_queue'
end
in
aux (List.fold_left f vertex_queue' edges)
in
aux (VertexSet.singleton (min_distance.(src), src));
min_distance, previous

let shortest_path_to (target : vertex) (previous : vertex array) : vertex list =
let rec aux target acc =
if target = -1 then
acc
else
aux previous.(target) (target :: acc)
in
aux target []

let adj_list =
[| [(1, 7.); (2, 9.); (5, 14.)]; (* 0 = a *)
[(0, 7.); (2, 10.); (3, 15.)]; (* 1 = b *)
[(0, 9.); (1, 10.); (3, 11.); (5, 2.)]; (* 2 = c *)
[(1, 15.); (2, 11.); (4, 6.)]; (* 3 = d *)
[(3, 6.); (5, 9.)]; (* 4 = e *)
[(0, 14.); (2, 2.); (4, 9.)] (* 5 = f *)
|]

let () =
let min_distance, previous = dijkstra 0 adj_list in
Printf.printf "Distance from 0 to 4: %f\n" min_distance.(4);
let path = shortest_path_to 4 previous in
print_string "Path: ";
List.iter (Printf.printf "%d, ") path;
print_newline ()</lang>


=={{header|PARI/GP}}==
=={{header|PARI/GP}}==