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 PriorityQueue
  def initialize
    @q = []
    @h = {}
  end

  # push obj unless already contained
  def push_ex(obj, cost)
    return if @h.key?(obj)
    i = (0...@q.size).bsearch{|i| @q[i][1] > cost} || @q.size
    @q.insert(i, [obj, cost])
    @h[obj] = true
  end

  def pop
    return nil if @q.empty?
    obj = @q.shift[0]
    @h.delete(obj)
    return obj
  end

  # - Node number starts from 1
  def self.dijkstra(edges, start)
    best = Array.new(edges.size+1){ [Float::INFINITY, nil] }
    best[start] = [0, [start]]
    q = PriorityQueue.new
    q.push_ex([start, [start]], 0)
    while (item = q.pop)
      a, route = *item
      edges[a].each do |b, cost|
        new_cost = best[a][0] + cost
        #p a: a, route: route, b: b, cost: cost, best: best[b][0], new_cost: new_cost
        if new_cost < best[b][0]
          best[b] = [new_cost, route+[b]]
          q.push_ex([b, route+[b]], new_cost)
        end
      end
    end
    return best
  end
end

N, M, *rest = $stdin.read.split.map(&:to_i)
EDGES = rest.each_slice(3).to_a
edges = Array.new(N+1){ [] }
EDGES.each do |a, b, cost|
  edges[a] << [b, cost]
  edges[b] << [a, cost]
end

used = Array.new(N+1){ Array.new(N+1) }
for i in 1..N
  best = PriorityQueue.dijkstra(edges, i)
  best.each do |_, route|
    next unless route
    route.each_cons(2) do |j, k|
      if j < k
        used[j][k] = true
      else
        used[k][j] = true
      end
    end
  end
end

p EDGES.count{|a, b| 
  a < b ? !used[a][b] : !used[b][a]
}