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: Accepted)

Scanf.scanf "%d %d" (fun n m ->
    let module S = Set.Make (struct type t = int let compare = compare end) in
    let mat = Array.make_matrix n n (max_int / 1000, S.empty) in
    for i = 0 to n - 1 do mat.(i).(i) <- 0,S.empty done;
    for i = 0 to m - 1 do
        Scanf.scanf " %d %d %d" (fun a b c ->
            let a = a - 1 in let b = b - 1 in
            mat.(a).(b) <- c, S.singleton i;
            mat.(b).(a) <- c, S.singleton i;
        )
    done;
    for k = 0 to n - 1 do
        for i = 0 to n - 1 do
            for j = 0 to n - 1 do
                let d1, s1 = mat.(i).(j) in
                let d2, s2 = mat.(i).(k) in
                let d3, s3 = mat.(k).(j) in
                if d1 > d2 + d3 then mat.(i).(j) <- d2 + d3, S.union s2 s3
            done
        done
    done;
    let edge = Array.make m false in
    for i = 0 to n - 1 do
        for j = 0 to n - 1 do
            let _, s = mat.(i).(j) in
            S.iter (fun v -> edge.(v) <- true) s
        done
    done;
    Array.fold_left (fun acc v -> if v then acc else acc + 1) 0 edge |> Printf.printf "%d\n"
)