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)

N,M = gets.split.map(&:to_i)
INF = M*1000*10
nodes = (1..N).to_a
es = {}
unused = {}
M.times {
  a,b,c = gets.split.map(&:to_i)
  a,b = b,a if a > b
  es[a] ||= {}; es[b] ||= {}
  es[a][b] = es[b][a] = c
  unused[(a-1)*N+(b-1)] = 1
}

def dijk(start, nodes, edges)  
  froms = nodes.map{ nil }
  ds = nodes.map{ INF }
  ds[start-1] = 0
  queue = [start]
  while f = queue.shift
    #p queue
    u = ds[f-1]
    edges[f].each{ |t,c|
      v = ds[t-1]
      next if u+c >= v
      ds[t-1] = u+c
      queue.push(t)
      froms[t-1] = f
    }
  end
  #p ds
  froms.map.with_index{ |f,i| f ? [f, i + 1].sort : nil }.compact
end

nodes.each{ |s|
	dijk(s, nodes, es).each{ |f,t|
      unused.delete((f-1)*N+(t-1))
    }
}

puts unused.size