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: Accepted)

class Adjmat
  
  def initialize(n=0, inf=9876543210)
    @n = n
    @inf = inf
    @mat = Array.new(n){ Array.new(n, inf ) }.each_with_index{|s,i| s[i] = 0 }
  end
  
  def add_edge(s,t,d=1)
    @mat[s][t] = d
    @mat[t][s] = d
  end
  
  def add_directed_edge(s,t,d=1)
    @mat[s][t] = d
  end
  
  def floyd_warshall!
    @n.times do |k|
      @n.times do |i|
        @n.times do |j|
          l = @mat[i][k]
          r = @mat[k][j]
          next if l == @inf || r == @inf
          nd = l + r
          @mat[i][j] = nd if @mat[i][j] > nd
        end
      end
    end
    @mat
  end
  
  def floyd_warshall
    res = @mat
    @n.times do |k|
      @n.times do |i|
        @n.times do |j|
          l = res[i][k]
          r = res[k][j]
          next if l == @inf || r == @inf
          res[i][j] = l + r if res[i][j] > l + r
        end
      end
    end
    res
  end
  
  def negative_cycle?
    @n.times{ |i| return true if @mat[i][i] < 0}
    false
  end
  
  def out
    @n.times do |i|
      @n.times do |j|
        print @mat[i][j] == @inf ? "INF" : @mat[i][j]
        print j != @n-1 ? " " : "\n"
      end
    end
  end
end

n, m = gets.to_s.split.map{|t| t.to_i }

a = Adjmat.new(n)

edges = Array.new(m){ gets.to_s.split.map{|t| t.to_i } }

edges.each do |e|
  s = e[0] - 1
  t = e[1] - 1
  c = e[2]
  a.add_edge(s,t,c)
end

a = a.floyd_warshall

ans = 0
edges.each do |e|
  s = e[0] - 1
  t = e[1] - 1
  c = e[2]
  ok = true
  n.times do |i|
    n.times do |j|
      if a[i][j] == a[i][s] + c + a[t][j]
        ok = false
        break
      end
    end
  end
  ans += 1 if ok
end

puts ans