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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

N, M = gets.split.map(&:to_i)
path = Array.new(N+1) { Hash.new }
abc = M.times.map do
  a, b, c = gets.split.map(&:to_i)
  path[a][b] = c
  path[b][a] = c
  [a, b, c]
end
d = Array.new(N+1) { Array.new(N+1, nil) }
(1..N).each { |i| d[i][i] = 0 }

(1..N).each do |s|
  ((s+1)..N).each do |g|
    next if d[s][g]
    a = [[[s], 0]]
    goal = nil
    min = 1_000_000
    a.each do |v, k|
      path[v.last].each do |(n, kk)|
        next if kk + k > min
        next if v.include?(n)
        if n == g
          min = k + kk
          goal = v + [n]
        elsif d[n][g] && k + kk + d[n][g] < min
          min = k + kk + d[n][g]
          goal = v + [n, g]
        else
          a << [v + [n], k + kk]
        end
      end
    end
    (goal.size - 1).times do |i|
      z = 0
      goal[i..-1].each_cons(2) do |j, k|
        z += d[j][k] || path[j][k]
        d[goal[i]][k] = z
        d[k][goal[i]] = z
      end
    end
  end
end

p(abc.count { |a,b,c| d[a][b] < c })