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;

class D {
    class Graph {
        public int[,] A;

        public Graph(int n) {
            A = new int[n, n];
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < n; ++j)
                    A[i, j] = int.MaxValue >> 1;
        }

        public Graph(Graph g) { A = g.A.Clone() as int[,]; }

        public void AddEdge(int s, int t, int c) {
            A[s, t] = A[t, s] = c;
        }

        public void Floyd() {
            for (int i = 0; i < A.GetLength(0); ++i)
                for (int j = 0; j < A.GetLength(0); ++j)
                    for (int k = 0; k < A.GetLength(0); ++k) {
                        var newA = A[j, i] + A[i, k];
                        if (newA < A[j, k]) A[j, k] = newA;
                    }
        }
    }

    public static void Main() {
        var line = Console.ReadLine().Split();
        var n = int.Parse(line[0]);
        var m = int.Parse(line[1]);
        var graph = new Graph(n);
        for (int i = 0; i < m; ++i) {
            line = Console.ReadLine().Split();
            var a = int.Parse(line[0]) - 1;
            var b = int.Parse(line[1]) - 1;
            var c = int.Parse(line[2]);
            graph.AddEdge(a, b, c);
        }
        var graph1 = new Graph(graph);
        graph.Floyd();
        var answer = 0L;
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                if (graph1.A[i, j] != int.MaxValue >> 1)
                    if (graph.A[i, j] < graph1.A[i, j]) ++answer;
        Console.WriteLine(answer >> 1);
    }
}