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.Linq;
public class Program {
	public static void Main()
	{
		var list = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToList();
		
		var N = list[0];
		var M = list[1];
		int[] a,b,c;
		a = new int[M];
		b = new int[M];
		c = new int[M];
		
		// 全ての辺を配列にセット
		for(int i = 0; i < M; i++)
		{
			var d = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToList();
			a[i] = d[0] - 1;
			b[i] = d[1] - 1;
			c[i] = d[2];
		}
		
		const int MAX = 1000;
		int[,] dist = new int[N, N];
		for(int i = 0; i < N; i++)
		{
			for(int j = 0; j < N; j++)
			{
				dist[i, j] = MAX;
			}
			dist[i,i] = 0;
		}		
		for(int i = 0; i < M; i++)
		{
			dist[a[i], b[i]] = c[i];
			dist[b[i], a[i]] = c[i];
		}		

		for(int k = 0; k < N; k++) //中間点
		{
			for(int i = 0; i < N; i++) //始点
			{
				for(int j = 0; j < N; j++)　//終点
				{
					dist[i, j] = Math.Min(dist[i,j], dist[i,k] + dist[k, j]);
					dist[j, i] = Math.Min(dist[i,j], dist[i,k] + dist[k, j]);
				}
			}		
		}
		int ans = 0;
		for (var i = 0; i < M; i++)
		{
			if(dist[a[i], b[i]] < c[i]) 
				ans++;
		}
		Console.WriteLine(ans);
	}
}