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 Scala (2.13.1) (judge verdict: Accepted)

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n, m = sc.nextInt()
  val e = List.fill(m)((sc.nextInt()-1, sc.nextInt()-1, sc.nextInt()))

  def shortestPath(n: Int, edges: Seq[(Int, Int, Int)]): Array[Array[Int]] = {
    val ans = Array.tabulate(n,n)((i,j) => if (i == j) 0 else -1)
    for ((u, v, l) <- edges) {
      ans(u)(v) = l
      ans(v)(u) = l
    }

    for {
      k <- 0 until n
      i <- 0 until n
      j <- 0 until n
      if ans(i)(k) != -1 && ans(k)(j) != -1 &&
        (ans(i)(j) == -1 || ans(i)(k) + ans(k)(j) < ans(i)(j))
    } {
      ans(i)(j) = ans(i)(k) + ans(k)(j)
    }
    ans
  }

  val sp = shortestPath(n, e)
  val ans = e.count{ case (a, b, c) =>
    sp(a)(b) < c
  }
  println(ans)
}