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) { [] }
abc = M.times.map do
  a, b, c = gets.split.map(&:to_i)
  path[a] << [b, c]
  path[b] << [a, c]
  [a, b]
end
used = []
checked = Array.new(N+1) { Array.new(N+1, false) }

(1..(N-1)).each do |s|
  ((s+1)..N).each do |g|
    next if checked[s][g]
    goals = []
    a = [[[s], 0]]
    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
          goals << [v + [n], k + kk]
        end
        a << [v + [n], k + kk]
      end
    end
    v , _ = goals.min_by { |(_, k)| k }
    v.each_cons(2) do |i, j|
      used << [i, j]
    end
    v.each do |i|
      v.each do |j|
        i, j = j, i if i > j
        checked[i][j] = true
      end
    end
  end
end

used.each do |i,j|
  abc.delete([i,j])
  abc.delete([j,i])
end

p abc.size