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 (2.11.7) (judge verdict: Accepted)

import java.util.Scanner

object Main extends App{
  def warshall_floyd(d: Array[Array[Int]], v: Int) =
    for(k <- 1 to v; i <- 1 to v; j <- 1 to v){
      d(i)(j) = Math.min(d(i)(j), d(i)(k) + d(k)(j))
  }

  // input
  val sc = new Scanner(System.in)

  val n = sc.nextInt()
  val m = sc.nextInt()

  val graph = Array.ofDim[Int](n+1, n+1)
  val edges = new Array[(Int, Int, Int)](m+1)
  for(i <- 1 to n; j <- 1 to n)
    if(i == j)
      graph(i)(j) = 0
    else
      graph(i)(j) = 10000000

  for(i <- 1 to m){
    val a = sc.nextInt()
    val b = sc.nextInt()
    val c = sc.nextInt()

    edges(i) = (a, b, c)
    graph(a)(b) = c
    graph(b)(a) = c
  }


  // main process
  warshall_floyd(graph, n)

  var ans = 0
  for((a, b, c) <- edges){
    var flag = false
    for(i <- 1 to n; j <- 1 to n){
      if(graph(i)(a) + c + graph(b)(j) == graph(i)(j)
        || graph(i)(b) + c + graph(a)(j) == graph(i)(j))
        flag = true
    }

    if(!flag)
      ans += 1
  }

  println(ans)
}