結果

問題 No.1 道のショートカット
ユーザー vjudge1
提出日時 2025-09-05 21:00:11
言語 OCaml
(5.2.1)
結果
AC  
実行時間 5 ms / 5,000 ms
コード長 2,693 bytes
コンパイル時間 2,150 ms
コンパイル使用メモリ 22,768 KB
実行使用メモリ 7,716 KB
最終ジャッジ日時 2025-09-05 21:00:15
合計ジャッジ時間 3,233 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 4
other AC * 40
権限があれば一括ダウンロードができます

ソースコード

diff # 

let () =
  (* Read number of nodes, budget, and number of edges *)
  let num_nodes, budget, num_edges = Scanf.scanf "%d %d %d " (fun a b c -> (a, b, c)) in
  
  (* Arrays to store edge information *)
  let source_nodes = Array.make num_edges 0 in
  let target_nodes = Array.make num_edges 0 in
  let edge_costs = Array.make num_edges 0 in
  let edge_times = Array.make num_edges 0 in
  
  (* Read source nodes (convert to 0-based indexing) *)
  for i = 0 to num_edges - 1 do
    source_nodes.(i) <- (Scanf.scanf "%d " (fun x -> x)) - 1
  done;
  
  (* Read target nodes (convert to 0-based indexing) *)
  for i = 0 to num_edges - 1 do
    target_nodes.(i) <- (Scanf.scanf "%d " (fun x -> x)) - 1
  done;
  
  (* Read edge costs *)
  for i = 0 to num_edges - 1 do
    edge_costs.(i) <- Scanf.scanf "%d " (fun x -> x)
  done;
  
  (* Read edge times *)
  for i = 0 to num_edges - 1 do
    edge_times.(i) <- Scanf.scanf "%d " (fun x -> x)
  done;
  
  (* Create adjacency list for the graph *)
  let graph = Array.make num_nodes [] in
  for i = 0 to num_edges - 1 do
    let source = source_nodes.(i) in
    let target = target_nodes.(i) in
    let cost = edge_costs.(i) in
    let time = edge_times.(i) in
    graph.(source) <- (target, cost, time) :: graph.(source)
  done;
  
  (* Initialize DP table *)
  let infinity = 1000000000 in
  let dp = Array.make_matrix num_nodes (budget + 1) infinity in
  
  (* Start at node 0 with full budget, time = 0 *)
  dp.(0).(budget) <- 0;
  
  (* Dynamic programming: for each node, process all possible budget states *)
  for current_node = 0 to num_nodes - 1 do
    for remaining_budget = budget downto 0 do
      (* Skip if this state is unreachable *)
      if dp.(current_node).(remaining_budget) <> infinity then
        (* Process all outgoing edges from current node *)
        List.iter (fun (target, cost, time) ->
          (* Check if we have enough budget to take this edge *)
          if remaining_budget >= cost then
            let new_budget = remaining_budget - cost in
            let new_time = dp.(current_node).(remaining_budget) + time in
            (* Update if we found a better path *)
            if new_time < dp.(target).(new_budget) then
              dp.(target).(new_budget) <- new_time
        ) graph.(current_node)
    done
  done;
  
  (* Find the minimum time to reach the last node (node n-1) *)
  let min_time = ref infinity in
  for remaining_budget = 0 to budget do
    if dp.(num_nodes - 1).(remaining_budget) < !min_time then
      min_time := dp.(num_nodes - 1).(remaining_budget)
  done;
  
  (* Output result *)
  if !min_time = infinity then
    print_endline "-1"
  else
    Printf.printf "%d\n" !min_time
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