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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
  val (N, M) = readLine()!!.split(" ").map(String::toInt)

  val edges = mutableListOf<List<Int>>()
  (0 until M).forEach {
    val (a, b, c) = readLine()!!.split(" ").map(String::toInt)
    edges.add(listOf(a-1, b-1, c-1))
  }

  val MAX = 999999999
  val dists = Array(N, { Array<Int>(N, { MAX }) })
  (0 until N).forEach { dists[it][it] = 0 }

  edges.forEach { edge ->
    dists[edge[0]][edge[1]] = Math.min(dists[edge[0]][edge[1]], edge[2])
    dists[edge[1]][edge[0]] = Math.min(dists[edge[1]][edge[0]], edge[2])
  }

  (0 until N).forEach { k ->
    (0 until N).forEach { i -> 
      (0 until N).forEach { j ->
        dists[i][j] = Math.min(dists[i][j], dists[i][k] + dists[k][j])
      }
    }    
  }

  var ans = 0
  edges.forEach { edge ->
    if (dists[edge[0]][edge[1]] < edge[2]) {
      ans = ans + 1
    }
  }

  println(ans)
}