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.Numerics;
class Program
{     
    const int INF = short.MaxValue;
    public void Slove()
    {
        var inputs = Console.ReadLine().Split();
        int n = int.Parse(inputs[0]);
        int m = int.Parse(inputs[1]);
        int[][] distance =
            Enumerable.Range(0, n)
            .Select(x => new int[n].Select(y => INF).ToArray())
            .ToArray();
        foreach(int i in Enumerable.Range(0, n))
            distance[i][i] = 0;
        var node =
            Enumerable.Range(0, m)
            .Select(x => Console.ReadLine().Split()
                .Select(int.Parse).ToArray())
            .Select(x => new {a = x[0] - 1, b = x[1] - 1, dis = x[2]})
            .ToArray();
        foreach(var item in node)
        {
            distance[item.a][item.b] = item.dis;
            distance[item.b][item.a] = item.dis;
        }
        foreach(int i in Enumerable.Range(0, n))
            foreach(int j in Enumerable.Range(0, n))
                foreach(int k in Enumerable.Range(0, n))
                {
                    distance[j][k] = Math.Min(distance[j][k], distance[j][i] + distance[i][k]);
                }
        //System.Console.WriteLine(string.Join("\n", distance.Select(x => string.Join(",", x))));
        int result = node.Count(x => x.dis != distance[x.a][x.b]);
        System.Console.WriteLine(result);
    }
    
    static void Main(string[] args)
    {
        new Program().Slove();
    }
}