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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using static Input;

public class Prog
{
    private const int INF = 1000000007;
    public static void Main()
    {
        int N = NextInt();
        int M = NextInt();
        List<int>[] wf = new List<int>[N];
        for (int i = 0; i < N; i++) wf[i] = Enumerable.Repeat(INF, N).ToList();
        for (int i = 0; i < N; i++) wf[i][i] = 0;
        List<int>[] e = new List<int>[N];
        List<bool>[] e_b = new List<bool>[N];
        for (int i = 0; i < N; i++) e[i] = Enumerable.Repeat(2 * INF, N).ToList();
        for (int i = 0; i < N; i++) e_b[i] = Enumerable.Repeat(false, N).ToList();
        for (int i = 0; i < M; i++)
        {
            int A = NextInt() - 1;
            int B = NextInt() - 1;
            int Cost = NextInt();
            wf[A][B] = Cost;
            wf[B][A] = Cost;
            e[A][B] = Cost;
            e[B][A] = Cost;
            e_b[A][B] = true;
            e_b[B][A] = true;
        }

        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                for (int k = 0; k < N; k++)
                    wf[j][k] = Math.Min(wf[j][k], wf[j][i] + wf[i][k]);

        for (int s = 0; s < N; s++)
        {
            //始点
            for (int i = 0; i < N; i++)
            {
                if (s == i) continue;
                //経由点
                for (int j = 0; j < N; j++)
                {
                    //終点
                    if (s == j) continue;
                    if (!e_b[i][j]) continue;
                    if (wf[s][i] + e[i][j] == wf[s][j])
                    {
                        e_b[i][j] = false;
                        e_b[j][i] = false;
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < N; i++) ans += e_b[i].Count(v => v == true);
        Console.WriteLine(ans / 2);
    }
}


public class PriorityQueue<Tkey, Tvalue> where Tkey : IComparable
{
    private static List<Tkey> list = new List<Tkey>();
    private static List<Tvalue> value = new List<Tvalue>();
    public int Count { get; private set; } = 0;
    private static bool Descending;
    public PriorityQueue(bool b = true) { Descending = b; }
    private static int Compare(Tkey a, Tkey b)
    {
        if (Descending) return a.CompareTo(b);
        else return b.CompareTo(a);
    }
    public void Enqueue(Tkey e, Tvalue n)
    {
        Count++;
        int now = Count - 1;
        list.Add(e);
        value.Add(n);
        while (now != 0)
        {
            int par = (now - 1) / 2;
            if (Compare(list[par], e) <= 0) break;
            Tkey a = list[now]; list[now] = list[par]; list[par] = a;
            Tvalue b = value[now]; value[now] = value[par]; value[par] = b;
            now = par;
        }
    }
    public Tvalue Dequeue()
    {
        if (Count == 0) return (Tvalue)(object)(-1);
        Tvalue e = value.First();
        if (Count == 1) { list = new List<Tkey>(); value = new List<Tvalue>(); Count--; return e; }
        Tkey x = list[Count - 1];
        Tvalue y = value[Count - 1];
        int i = 0;
        while (i * 2 + 1 < Count)
        {
            int l = i * 2 + 1;
            int r = i * 2 + 2;
            if (r < Count && Compare(list[r], list[l]) < 0) l = r;
            if (Compare(list[l], x) >= 0) break;
            list[i] = list[l];
            value[i] = value[l];
            i = l;
        }
        list[i] = x;
        value[i] = y;
        Count--;
        return e;
    }
}

public class Input
{
    private static Queue<string> q = new Queue<string>();
    private static void Confirm() { if (q.Count == 0) foreach (var s in Console.ReadLine().Split(' ')) q.Enqueue(s); }
    public static int NextInt() { Confirm(); return int.Parse(q.Dequeue()); }
    public static long NextLong() { Confirm(); return long.Parse(q.Dequeue()); }
    public static string NextString() { Confirm(); return q.Dequeue(); }
    public static double NextDouble() { Confirm(); return double.Parse(q.Dequeue()); }
    public static int[] LineInt() { return Console.ReadLine().Split(' ').Select(int.Parse).ToArray(); }
    public static long[] LineLong() { return Console.ReadLine().Split(' ').Select(long.Parse).ToArray(); }
    public static string[] LineString() { return Console.ReadLine().Split(' ').ToArray(); }
    public static double[] LineDouble() { return Console.ReadLine().Split(' ').Select(double.Parse).ToArray(); }
}