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)

{-# OPTIONS_GHC -O #-}

import Control.Monad
import Control.Monad.ST
import Data.List
import Control.Monad.Trans.State.Strict
import Data.Char (isSpace)
import Data.Maybe (fromJust,fromMaybe)
import qualified Data.ByteString.Lazy.Char8 as B

import Data.Array.ST
import Data.Array.Unboxed
type Dist s = STUArray s (Int,Int) Int

type Parser a = StateT B.ByteString Maybe a

parseInt :: Parser Int
parseInt = StateT (B.readInt . B.dropWhile isSpace)

parseInt2 :: Parser (Int,Int)
parseInt2 = liftM2 (,) parseInt parseInt

parseInt3 :: Parser (Int,Int,Int)
parseInt3 = liftM3 (,,) parseInt parseInt parseInt

parse :: Parser ((Int,Int), [(Int,Int,Int)])
parse = do
    (n,m)<-parseInt2
    rs<-replicateM m parseInt3
    return ((n,m),rs)

main=do
    ((n,m),rs)<-fromJust.evalStateT parse <$> B.getContents
    let da = wf n rs :: UArray (Int,Int) Int
    --
    print $ length $ filter (f n da) rs

f n da (a,b,c) = any (\s->s1 s /= s2 s) vs
    where
        vs = [v|v<-[1..n],v/=a,v/=b]
        s1 s = max (da ! (s,b)) (da ! (s,a))
        s2 s =min (da ! (s,a) + da ! (a,b)) (da !(s,b) + da !(b,a))
--
wf::Int->[(Int,Int,Int)]->UArray (Int,Int) Int
wf n rs = runST $ do
    d <- newArray ((1,1),(n,n)) (maxBound`div`2) :: ST s (STUArray s (Int,Int) Int)
    forM_ rs $ \(u,v,l) -> writeArray d (u,v) l >> writeArray d (v,u) l
    forM_ [(k,i,j)|k<-[1..n],i<-[1..n],j<-[1..n]] $ \(k,i,j) -> do
        dc <- readArray d (i,j)
        dn <- (+) <$> readArray d (i,k) <*> readArray d (k,j)
        when (dc > dn) $ writeArray d (i,j) dn
    freeze d