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 System.Text;
using System.IO;
using System.Globalization;
using static System.Console;
using Pair = System.Collections.Generic.KeyValuePair<int, int>;

class Program
{
    static void Main()
    {
        var sw = new StreamWriter(OpenStandardOutput()) { AutoFlush = false };
        SetOut(sw);
        new Program().solve();
        Out.Flush();
    }
    Scanner cin = new Scanner();
    readonly int[] dd = { 0, 1, 0, -1, 0 };
    readonly int mod = 1000000007;
    readonly string alfa = "abcdefghijklmnopqrstuvwxyz";

    struct Edge
    {
        public int u, v, cost,check;
        public Edge(int u, int v, int cost,int check)
        {
            this.u = u;
            this.v = v;
            this.cost = cost;
            this.check = check;
        }
    }

    int V, E;

    void solve()
    {
        V = cin.nextint;
        E = cin.nextint;

        var D = new int[V, V];
        for (int i = 0; i < V; i++)
        {
            for (int j = 0; j < V; j++)
            {
                D[i, j] = (i == j) ? 0 : int.MaxValue / 3 ;
            }

        }
        var G = new Edge[E];
        for (int i = 0; i < E; i++)
        {
            var t = cin.scanint;
            G[i] = new Edge(t[0] - 1, t[1] - 1, t[2], 0);
            D[t[0] - 1, t[1] - 1] = t[2];
            D[t[1] - 1, t[0] - 1] = t[2];
        }

        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]);
                }
            }
        }

        for (int k = 0; k < V; k++)
        {
            for (int i = 0; i < V; i++)
            {
                for (int j = 0; j < E; j++)
                {
                    if (D[k, i] == G[j].cost && ((k == G[j].v && i == G[j].u) || (k == G[j].u && i == G[j].v)))
                    {
                        G[j].check = 1;
                        
                    }
                }
            }
        }
        int cnt = 0;
        for (int i = 0; i < E; i++)
        {
            cnt += G[i].check;
        }
        //Console.WriteLine(D[1,2]);
        WriteLine(E - cnt);
    }

}

class Scanner
{
    string[] s; int i;
    char[] cs = new char[] { ' ' };
    public Scanner() { s = new string[0]; i = 0; }
    public string[] scan { get { return ReadLine().Split(); } }
    public int[] scanint { get { return Array.ConvertAll(scan, int.Parse); } }
    public long[] scanlong { get { return Array.ConvertAll(scan, long.Parse); } }
    public double[] scandouble { get { return Array.ConvertAll(scan, double.Parse); } }
    public string next
    {
        get
        {
            if (i < s.Length) return s[i++];
            string st = ReadLine();
            while (st == "") st = ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next;
        }
    }
    public int nextint { get { return int.Parse(next); } }
    public long nextlong { get { return long.Parse(next); } }
    public double nextdouble { get { return double.Parse(next); } }
}