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 Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import qualified Data.Set as S
import qualified Data.ByteString.Char8 as B
import qualified Data.Array.IArray as A
import qualified Data.Map.Strict as M

main = do
  [n,m] <- map read . words <$> getLine
  es <- readUndirectedEdge <$> B.getContents
  print (nsp n m es)

nsp n m es = m - (foldr (check n memo) 0 es)
  where
    memo = warshallFloyd n es

check n memo ((i,j),w) acc
  | any (\x -> memo !? (x,j) - memo !? (x,i) == w) [1..n] = acc + 1
  | otherwise = acc

(!?) m k = M.findWithDefault 100000 k m

type Memo = M.Map (Vertex, Vertex) Weight

warshallFloyd :: Int -> [Edge] -> Memo
warshallFloyd n es = foldl shorten m0 kij
  where
    m0 = M.fromList es
    kij = [(k, i, j) | k <- [1 .. n], i <- [1 .. n], j <- [1 .. n]]

shorten :: Memo -> (Vertex, Vertex, Vertex) -> Memo
shorten m (k, i, j)
  | i == j = m
  | otherwise = case connect m k i j of
    Nothing -> m
    Just w -> M.insertWith min (i, j) w m

connect :: Memo -> Vertex -> Vertex -> Vertex -> Maybe Weight
connect m k i j = do
  w1 <- M.lookup (i, k) m
  w2 <- M.lookup (k, j) m
  return (w1 + w2)

type Bound = (Vertex, Vertex)
type Vertex = Int
type Weight = Int
type Edge = ((Vertex, Vertex), Weight)
type Path = [Vertex]
type Graph = A.Array Vertex (S.Set (Vertex, Weight))

buildG :: Bound -> [Edge] -> Graph
buildG b = A.accumArray (flip S.insert) S.empty b . map (\((s, t), w) -> (s, (t, w)))

from :: Graph -> Vertex -> S.Set (Vertex, Weight)
from = (A.!)

size :: Graph -> Int
size g = let (i,j) = A.bounds g in j - i + 1

readUndirectedEdge :: B.ByteString -> [Edge]
readUndirectedEdge = concatMap (constructTwoWay . map readInt . B.words) . B.lines

constructTwoWay :: [Int] -> [Edge]
constructTwoWay [s,t,w] = [((s, t), w), ((t, s), w)]
constructTwoWay _ = undefined

readInt :: B.ByteString -> Int
readInt = maybe undefined fst . B.readInt