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: Accepted)

fun main(arg: Array<String>) {
    val (N, M) = listOfInt()
    val G = Array(N) { IntArray(N) { 100000007 } }
    val E = Array(N) { IntArray(N) }
    repeat(N) { n -> G[n][n] = 0 }
    repeat(M) {
        val (a, b, c) = listOfInt(-1, -1, 0)
        G[a][b] = c; G[b][a] = c
        E[a][b] = c; E[b][a] = c
    }
    for (k in 0 until N)
    for (n in 0 until N)
    for (m in 0 until N)
        G[n][m] = Math.min(G[n][m], G[n][k] + G[k][m])
    val A = Array(N) { IntArray(N) }
    for (s in 0 until N)
    for (t in 0 until N)
    for (n in 0 until N)
    for (m in 0 until N) {
        if (s == t) continue
        if (n == m) continue
        if (G[s][t] == G[s][n] + E[n][m] + G[m][t]) A[n][m] += 1
    }
    var ans = 0
    for (n in 0 until N)
        for (m in 0 until N) {
            if (A[n][m] == 0 && E[n][m] != 0) ans += 1
        }
    println(ans / 2)
}

fun next() = readLine()!!
fun nextInt(offset: Int = 0) = Integer.parseInt(next()) + offset
fun nextLong() = next().toLong()
fun listOfString() = next().split(" ")
fun listOfInt() = listOfString().map(String::toInt)
fun listOfInt(offset: Int = 0) = listOfString().map { Integer.parseInt(it) + offset }
fun listOfLong() = listOfString().map(String::toLong)
fun tripleIntIntLong() = listOfLong().let { Triple(it[0].toInt(), it[1].toInt(), it[2]) }
fun listOfInt(vararg offset: Int) = listOfString().map { Integer.parseInt(it) }.mapIndexed { index, int -> int + offset[index % offset.size] }