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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using static System.Console;
using System.Runtime.CompilerServices;
using static MyUtil;

class MyUtil
{
    public static int[] ReadIntArray()
    {
	return ReadLine().Split().Select(x => int.Parse(x)).ToArray();
    }

    public static void WarshallFloyd(int[,] d, int v)
    {
	for (int k = 0; k < v; k++)
	    for (int i = 0; i < v; i++)
		for (int j = 0; j < v; j++)
		    d[i, j] = Math.Min(d[i, j], d[i, k] + d[k, j]);
    }
}

class Program
{
    public static void Main()
    {
	int M = 1000000;
	var tmp = ReadIntArray();
	int n = tmp[0], m = tmp[1];
	int[,] d = new int[n,n];

	for (int i = 0; i < n; i++)
	{
	    for (int j = 0; j < n; j++)
	    {
		if (i==j) d[i,j] = 0;
		else      d[i,j] = M;
	    }
	}

	List<Tuple<int,int,int>> e = new List<Tuple<int,int,int>>();
	for (int i = 0; i < m; i++)
	{
	    tmp = ReadIntArray();
	    int a = tmp[0], b = tmp[1], c = tmp[2];
	    e.Add(Tuple.Create(a-1, b-1, c));
	    d[a-1, b-1] = c;
	    d[b-1, a-1] = c;
	}

	WarshallFloyd(d, n);
	
	int ans = m;
	
	foreach (Tuple<int,int,int> t in e)
	{
	    int a = t.Item1, b = t.Item2, c = t.Item3;
	    for (int j = 0; j < n; j++)
	    {
		if (d[j,a] + c == d[j,b])
		{
		    ans--;
		    break;
		}
	    }
	}
	
	WriteLine(ans);
    }
}