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 LuaJIT (2.0.4) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local edge = {}
for i = 1, n do edge[i] = {} end
local edgeinfo = {}
for i = 1, m do
  local a, b, c = io.read("*n", "*n", "*n")
  edge[a][b], edge[b][a] = c, c
  table.insert(edgeinfo, {a, b, c})
end

local function getlength(start_idx)
  local taskstate = {}
  for i = 1, n do taskstate[i] = false end
  local tasks = {}
  local tasknum = 0
  local done = 0

  local len = {}
  for i = 1, n do len[i] = 100000 end
  len[start_idx] = 0

  local function addtask(idx)
    if not taskstate[idx] then
      taskstate[idx] = true
      tasknum = tasknum + 1
      tasks[tasknum] = idx
    end
  end
  addtask(start_idx)

  while done < tasknum do
    done = done + 1
    local src = tasks[done]
    taskstate[src] = false
    local tmp = {}
    for dst, cost in pairs(edge[src]) do
      if len[src] + cost < len[dst] then
        len[dst] = len[src] + cost
        table.insert(tmp, dst)
      end
    end
    table.sort(tmp, function(a, b) return len[a] < len[b] end)
    for i = 1, #tmp do addtask(tmp[i]) end
  end
  return len
end
local lenall = {}
for i = 1, n do
  lenall[i] = getlength(i)
end
local ret = 0
for i = 1, m do
  local contains = false
  local edge_l, edge_r, c = edgeinfo[i][1], edgeinfo[i][2], edgeinfo[i][3]
  for src = 1, n do
    for dst = 1, n do
      if lenall[src][dst] == lenall[src][edge_l] + c + lenall[edge_r][dst] then
        contains = true
        break
      end
    end
    if contains then break end
  end
  if not contains then ret = ret + 1 end
end
print(ret)