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;

namespace D___Candidates_of_No_Shortest_Paths
{
    class Program
    {
        static void Main(string[] args)
        {
            var read = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
            var N = read[0];
            var M = read[1];

            //2次元配列に辺と重さを記録
            var edges = new int[M,3];
            for(int i = 0 ; i < M ; i++)
            {
                read = Array.ConvertAll(Console.ReadLine().Split(' '),int.Parse);
                edges[i,0] = read[0];
                edges[i,1] = read[1];
                edges[i,2] = read[2];
            }

            //2次元配列に各頂点間の最短距離を記録
            var costs = new int[N+1,N+1];
            for(int i = 1 ; i < N+1 ; i++) for(int j = 1 ; j < N+1 ; j++) costs[i,j] = 1000000000;
            for(int i = 1 ; i < N+1 ; i++) costs[i,i] = 0;

            for(int i = 0 ; i < M ; i++)
            {
                costs[edges[i,0],edges[i,1]] = edges[i,2];
                costs[edges[i,1],edges[i,0]] = edges[i,2];
            }

            //Warshall-Floyd法で全ての頂点間の最短距離求める
            for(int via = 1 ; via < N+1 ; via++)
                for(int from = 1 ; from < N+1 ; from++)
                    for(int to = 1 ; to < N+1 ; to++) 
                        costs[from,to] = Math.Min(costs[from,to],costs[from,via]+costs[via,to]);

            //各辺の距離と両端の頂点間の最短距離を比較し、辺の距離のほうが長いものの数を出力
            var result = 0;
            for(int i = 0 ; i < M ; i++) if(costs[edges[i,0],edges[i,1]] != edges[i,2]) result++;

            Console.WriteLine(result);
            

        }
    }
}