Problem:
Score : 400 points

Problem Statement

You are given an undirected connected weighted graph with N vertices and M edges that contains neither self-loops nor double edges.

The i-th (1≤i≤M) edge connects vertex a_i and vertex b_i with a distance of c_i.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

A connected graph is a graph where there is a path between every pair of different vertices.

Find the number of the edges that are not contained in any shortest path between any pair of different vertices.

Constraints

2≤N≤100

N-1≤M≤min(N(N-1)/2,1000)

1≤a_i,b_i≤N

1≤c_i≤1000

c_i is an integer.

The given graph contains neither self-loops nor double edges.

The given graph is connected.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the number of the edges in the graph that are not contained in any shortest path between any pair of different vertices.

Sample Input 1

3 3
1 2 1
1 3 1
2 3 3

Sample Output 1

1

In the given graph, the shortest paths between all pairs of different vertices are as follows:

The shortest path from vertex 1 to vertex 2 is: vertex 1 → vertex 2, with the length of 1.

The shortest path from vertex 1 to vertex 3 is: vertex 1 → vertex 3, with the length of 1.

The shortest path from vertex 2 to vertex 1 is: vertex 2 → vertex 1, with the length of 1.

The shortest path from vertex 2 to vertex 3 is: vertex 2 → vertex 1 → vertex 3, with the length of 2.

The shortest path from vertex 3 to vertex 1 is: vertex 3 → vertex 1, with the length of 1.

The shortest path from vertex 3 to vertex 2 is: vertex 3 → vertex 1 → vertex 2, with the length of 2.

Thus, the only edge that is not contained in any shortest path, is the edge of length 3 connecting vertex 2 and vertex 3, hence the output should be 1.

Sample Input 2

3 2
1 2 1
2 3 1

Sample Output 2

0

Every edge is contained in some shortest path between some pair of different vertices.

Solution in OCaml (4.02.3) (judge verdict: Time Limit Exceeded)

type 'a set =
  | Empty
  | Singleton of 'a
  | Union of 'a set * 'a set

module IntSet = Set.Make (struct
  type t = int
  let compare = compare
end)

let rec make_intset acc = function
  | Empty -> acc
  | Singleton x -> IntSet.add x acc
  | Union (s, t) -> make_intset (make_intset acc s) t

let warshall_floyd ( + ) min n d =
  for i = 0 to n - 1 do
    for j = 0 to n - 1 do
      for k = 0 to n - 1 do
        d.(j).(k) <- min d.(j).(k) (d.(j).(i) + d.(i).(k))
      done
    done
  done

let () =
  let n, m = Scanf.scanf "%d %d\n" (fun n m -> n, m) in
  let dp = Array.make_matrix n n (infinity, Empty) in
  for i = 0 to n - 1 do
    dp.(i).(i) <- (0., Empty)
  done;
  for i = 0 to m - 1 do
    Scanf.scanf "%d %d %d\n" (fun a b c ->
      dp.(a - 1).(b - 1) <- (float_of_int c, Singleton i);
      dp.(b - 1).(a - 1) <- (float_of_int c, Singleton i))
  done;
  warshall_floyd
    (fun (d, es) (d', es') -> (d +. d', Union (es, es')))
    (fun (d, es) (d', es') ->
      match compare d d' with
      | -1 -> (d, es)
      | 0  -> (d, Union (es, es'))
      | 1  -> (d', es')) n dp;
  Array.map (fun dp ->
    Array.map snd dp
    |> Array.fold_left (fun s t -> Union (s, t)) Empty) dp
  |> Array.fold_left (fun s t -> Union (s, t)) Empty
  |> make_intset IntSet.empty
  |> IntSet.cardinal
  |> (fun n -> Printf.printf "%d\n" (m - n))