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)

# coding: utf-8
#
# :u ノードのインデックス
# :k u の出次数
# :v u に隣接する頂点の番号
# :w 隣接する頂点への重み
#
INF  = Float::INFINITY
Node = Struct.new(:u, :k, :v, :w)

class Graph
  attr :nodes, :dist
  def initialize(n)
    @nodes = Array.new(n){ Node.new }
    @nodes = @nodes.map.with_index do |node,idx|
      node.u = idx
      node
    end
    @dist = Array.new(n).map{Array.new(n, 0)}
  end

  def add_graph_edge(u, v, w)
    #puts "#{u} -> #{v} (#{w})"
    if @nodes[u].v.nil?
      @nodes[u].v = [v]
      @nodes[u].w = {v=>w}
    elsif @nodes[u].v.is_a?(Array)
      @nodes[u].v << v
      @nodes[u].w = {v=>w}.merge(@nodes[u].w)
    else
      @nodes[u].v = [@nodes[u].v, v]
      @nodes[u].w = {v=>w}.merge(@nodes[u].w)
    end
    @nodes[u].k = if @nodes[u].v.is_a?(Array)
                    @nodes[u].v.length
                  else
                    1
                  end
  end

  def warshall_floyd()
    # initialize
    for i in 0...@nodes.length
      for j in 0...@nodes.length
        @dist[i][j] = if not @nodes[i].w.nil? and @nodes[i].w.has_key?(j)
                        @nodes[i].w[j]
                      else
                        INF
                      end
      end
      @dist[i][i] = 0
    end

    # calc
    for k in 0...@nodes.length
      for i in 0...@nodes.length
        for j in 0...@nodes.length
          if @dist[i][k] != INF and @dist[k][j] != INF
            @dist[i][j] = [@dist[i][j], @dist[i][k] + @dist[k][j]].min
          end
        end
      end
    end
  end

  def has_negative_cycle?()
    ans = false
    for v in 0...@nodes.length
      ans = true if @dist[v][v] < 0
    end
    ans
  end
end

lines = $stdin.read
array = lines.split("\n")

N,M = array[0].split(" ").map(&:to_i)
graph = Graph.new(N)

array[1..M].each do |s|
  s,t,d = s.split(" ").map(&:to_i)
  graph.add_graph_edge(s-1,t-1,d)
  graph.add_graph_edge(t-1,s-1,d)
end

# execute WarshallFloyd !
graph.warshall_floyd()

ans = array[1..M].count do |s|
  s,t,d = s.split(" ").map(&:to_i)
  s,t = s-1,t-1
  graph.dist[s][t] != d
end

puts ans