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;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;

class Prog
{
    public Prog() { }
    public static int Main()
    {
        new Prog().calc();
        return 0;
    }

    Scanner cin;

    int[,] dist;
    int N;
    int M;
    const int INF = int.MaxValue / 2;


    public void calc()
    {
        cin = new Scanner();
        N = cin.nextInt();
        M = cin.nextInt();

        int[] a = new int[M];
        int[] b = new int[M];
        int[] c = new int[M];

        for(int i = 0;i<M;i++)
        {
            a[i] = cin.nextInt();
            b[i] = cin.nextInt();
            c[i] = cin.nextInt();
        }

        dist = new int[N + 1, N + 1];

        for (int i = 0; i < N + 1; i++)
        {
            for (int j = 0; j < N + 1; j++)
            {
                dist[i, j] = i == j ? 0 : INF;
            }
        }

        for(int i = 0;i<M;i++)
        {
            dist[a[i], b[i]] = c[i];
            dist[b[i], a[i]] = c[i];
        }

        //ワーシャルフロイドで全点間最短距離を求める
        for (int k = 1; k <= N; k++)
        {
            for (int i = 1; i <= N; i++)
            {
                for (int j = 1; j <= N; j++)
                {
                    dist[i, j] = Math.Min(dist[i, j], dist[i, k] + dist[k, j]);
                }
            }
        }

        int ans = 0;
        for(int i = 0;i<M;i++)
        {
            //ある辺の長さがその両端の頂点間の最短距離より長いなら、
            //その辺は（ほかの頂点間の最短経路としても）使われない
            if (dist[a[i], b[i]] < c[i]) ans++;
        }

        Console.WriteLine(ans);
    }

}


class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }

}