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;

public class Node
{
    public List<int> Edge = new List<int>();
    public List<int> EdgeCost = new List<int>();
    public bool Done = false;
    public int Cost = -1;
    public int Previous = -1;
    public int Id;

    public Node Clone()
    {
        return new Node()
        {
            Edge = this.Edge,
            EdgeCost = this.EdgeCost,
            Done = this.Done,
            Cost = this.Cost,
            Previous = this.Previous,
            Id = this.Id
        };
    }
}

public static partial class Program
{
    public static int N;
    public static Node[] Nodes;

    public static void Main()
    {
        var nm = CRL().ArrayLongs();
        N = (int) nm[0];
        Nodes = new Node[N];
        var m = nm[1];
        for (int i = 0; i < N; i++)
        {
            Nodes[i] = new Node()
            {
                Id = i
            };
        }

        for (var i = 0; i < m; i++)
        {
            var p = CRL().Split().Select(x => int.Parse(x) - 1).ToArray();
            Nodes[p[0]].Edge.Add(p[1]);
            Nodes[p[1]].Edge.Add(p[0]);
            Nodes[p[0]].EdgeCost.Add(p[2] + 1);
            Nodes[p[1]].EdgeCost.Add(p[2] + 1);
        }

        var memo = new List<KeyValuePair<int, int>>();
        for (int s = 0; s < N; s++)
        {
            for (int g = s + 1; g < N; g++)
            {
                var val = Dijkstra(s, g).ToArray();
                for (int i = 0; i < val.Length - 1; i++)
                {
                    memo.Add(new KeyValuePair<int, int>(val[i], val[i + 1]));
                    memo.Add(new KeyValuePair<int, int>(val[i + 1], val[i]));
                }
            }
        }

        var dist = memo.Distinct();
        CWL(m - dist.Count() / 2);
    }

    public static IEnumerable<int> Dijkstra(int start, int goal)
    {
        var nodes = new Node[N];
        for (int i = 0; i < N; i++)
        {
            nodes[i] = Nodes[i].Clone();
        }

        nodes[start].Cost = 0;

        while (true)
        {
            var done = new Node();
            var changed = false;
            foreach (var node in nodes)
            {
                if (node.Done || node.Cost < 0)
                {
                    continue;
                }

                if (changed == false || node.Cost < done.Cost)
                {
                    done = node;
                    changed = true;
                }
            }

            if (changed == false)
            {
                break;
            }

            done.Done = true;

            //更新
            for (int i = 0; i < done.Edge.Count; i++)
            {
                var next = done.Edge[i];
                var cost = done.Cost + done.EdgeCost[i];
                if (nodes[next].Cost < 0 || cost < nodes[next].Cost)
                {
                    nodes[next].Cost = cost;
                    nodes[next].Previous = done.Id;
                }
            }
        }

        var current = goal;
        while (current >= 0)
        {
            var next = nodes[current];
            yield return next.Id;
            current = next.Previous;
        }
    }
}

public static partial class Program
{
    //テンプレここから
    private static long ToInt(this string str) => long.Parse(str);
    private static long[] ToInt(this IEnumerable<string> strs) => strs.Select(x => x.ToInt()).ToArray();
    public static long[] ArrayLongs(this string str, char opr = ' ') => str.Split(opr).ToInt();
    public static string CRL() => Console.ReadLine();
    public static void CWL(object obj) => Console.WriteLine(obj);
    public static long Diff(long a, long b) => Math.Abs(Math.Max(a, b) - Math.Min(a, b));
    public static long Sign(this long i) => Math.Sign(i);
    public static bool IsEven(this long i) => i % 2 == 0;
    public static bool IsEven(this int i) => i % 2 == 0;
    public static long[] dx = new long[] {0, -1, 0, 1};
    public static long[] dy = new long[] {-1, 0, 1, 0};
    public static long mod = 1000000007;
    public static string alphabet = "abcdefghijklmnopqrstuvwxyz";
    public static long ToLong(this char c) => c.ToString().ToInt();

    public static long[] Sort(this long[] array)
    {
        var list = array.ToList();
        list.Sort();
        return list.ToArray();
    }

    public static long GCD(long a, long b) => b == 0 ? a : GCD(b, a % b);

    public static long LCM(long a, long b) => a * b / GCD(a, b);
    //テンプレここまで
}