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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.io.PrintWriter
import java.util.*
import java.math.*
import java.lang.*
import kotlin.comparisons.*

val pw = PrintWriter(System.out)
val MOD: Long = (1e+9 + 7).toLong()

data class Point(val next: Int, val cost: Int)
fun main(args: Array<String>) {
	val (n, m) = readListOfInt()
	val edges = Array(n) { mutableListOf<Point>() }
	for(i in 0 until m) {
		val (a, b, c) = readListOfInt().toMutableList().apply {
			this[0]--
			this[1]--
		}
		edges[a].add(Point(b, c))
		edges[b].add(Point(a, c))
	}

	val distances = Array(n) { Array(n) { 1000000 } }
	for(i in 0 until n) {
		distances[i][i] = 0
	}
	edges.forEachIndexed { a, nextEdges ->
		nextEdges.forEach { e ->
			val (b, c) = e
			distances[a][b] = c 
			distances[b][a] = c 
		}
	}
	for(k in 0 until n) {
		for(i in 0 until n) {
			for(j in 0 until n) {
				distances[i][j] = Math.min(distances[i][j], distances[i][k] + distances[k][j])
			}
		}
	}

	var cnt = m * 2
	edges.forEachIndexed { a, nextEdges ->
		nextEdges.forEach { e ->
			val (b, c) = e
			var isShortestPath = false
			for(i in 0 until n) {
				if(distances[i][a] + c == distances[i][b]) {
					isShortestPath = true	
				}
			}
			if(isShortestPath) {
				cnt--
			}
		}

	}
	println(cnt/2)
	pw.flush()
}

/****** Declared Functions and Data Structures ******/
// IO
fun read() = readLine()!!
fun readInt() = read().toInt()
fun readLong() = read().toLong()
fun readDouble() = read().toDouble()
fun readListOfString() = read().split(" ")
fun readListOfInt() = readListOfString().map { it.toInt() }
fun readListOfLong() = readListOfString().map { it.toLong() }
fun readListOfDouble() = readListOfString().map { it.toDouble() }

fun Double.format(digits: Int) = String.format("%.${digits}f", this)
fun Float.format(digits: Int) = String.format("%.${digits}f", this)

fun print(value: Any) { pw.print(value) }
fun println(value : Any) { pw.println(value) }

// Extensions
fun<T> List<T>.toBuckets(): Map<T, Int>
	= this.groupBy { it }.mapValues { it.value.size }

fun<T: Comparable<T>> List<T>.upperBound(element: T, fromIndex: Int=0, toIndex: Int=this.size): Int {
	var l = fromIndex
	var r = toIndex
	while(l < r) {
		val mid = (l+r) / 2
		if(element >= this[mid]) { 
			l = mid + 1
		} else {
			r = mid
		}
	}
	return l
}

fun<T: Comparable<T>> List<T>.lowerBound(element: T, fromIndex: Int=0, toIndex: Int=this.size): Int {
	var l = fromIndex
	var r = toIndex
	while(l < r) {
		val mid = (l+r) / 2
		if(element <= this[mid]) { 
			r = mid
		} else {
			l = mid + 1
		}
	}
	return r 
}

fun <K, V> Map<K, V>.toMutableMap(): MutableMap<K, V> = HashMap(this)

// Util Functions
fun Long.makeDivisors(): List<Long> {
	val divisors = mutableListOf<Long>()
	for(i in 1L .. Math.sqrt(this.toDouble()).toLong()+1L) {
		if(this % i == 0L) {
			divisors.add(i)
			if(i != this/i) {
				divisors.add(this/i)
			}
		}
	}
	return divisors.distinct().sorted()
}

fun Long.isPrime(): Boolean {
	if(this==1L) return false
	for(i in 2L .. Math.sqrt(this.toDouble()).toLong() + 1L) {
		if(this % i == 0L && this != i) {
			return false
		}
	}
	return true
}


fun<T> permutations(src: List<T>): List<List<T>> {
	if(src.size == 1) return listOf(src)
	val perms = mutableListOf<List<T>>()
	val insertElement = src[0]
	permutations(src.drop(1)).forEach { perm ->
		for(i in 0..perm.size) {
			val newPerm = perm.toMutableList()
			newPerm.add(i, insertElement)
			perms.add(newPerm.toList())
		}
	}
	return perms
}

// return nCr, if you want nCr, please access v[n][r]
fun combinations(n: Long): List<List<Long>> {
	val v = (0..n).map { (0..n).map{0L}.toMutableList() }.toMutableList()
	for(i in 0 until v.size) {
		v[i][0] = 1L
		v[i][i] = 1L
	}
	for(i in 0 until v.size) {
		for(j in 1 until i) {
			v[i][j] = (v[i-1][j-1] + v[i-1][j]) % MOD
		}
	}
	return v.map { it.toList() }.toList()
}

fun repeatedModPow(a: Long, n: Long, mod: Long): Long = 
	when {
		n == 1L -> a % mod
		n % 2L == 1L -> (a * repeatedModPow(a, n-1L, mod)) % mod
		else -> {
			val tmp = repeatedModPow(a, n / 2L, mod)
			(tmp * tmp) % mod
		}
	}

// return nCr % MOD
fun modnCkWithoutArray(n: Long, k: Long, mod: Long): Long {
	var ret = 1L
	val min = Math.min(k, n-k)
	// まず分母の計算
	for(i in n downTo (n - min + 1)) {
		ret = ret * i % mod
	}
	for(i in min downTo 1) {
		ret = ret * repeatedModPow(i, (mod - 2L), mod) % mod
	}
	return ret
}

// return nCr % MOD
fun modnCk(n: Int, k: Int, mod: Long): Long {
	val factorials = Array(n+1) { 0L }.apply {
		this[0] = 1L
		if(this.size > 1) this[1] = 1L
	}
	val inverses = Array(n+1) { 0L }.apply {
		this[0] = 1L
		if(this.size > 1) this[1] = 1L
	}
	val factorialInverses = Array(n+1) { 0L }.apply {
		this[0] = 1L
		if(this.size > 1) this[1] = 1L
	}
	for(i in 2 .. n) {
		factorials[i] = (factorials[i-1] * i.toLong() % mod)
		inverses[i] = (mod - inverses[mod.toInt() % i] * (mod / i.toLong()) % mod)
		factorialInverses[i] = (factorialInverses[i-1] * inverses[i] % mod)
	}

	return factorials[n] % mod * (factorialInverses[k] * factorialInverses[n-k] % mod) % mod
}

// should be a >= b
fun gcd(a: Long, b: Long): Long = 
	if(b == 0L) a else gcd(b, a % b)

// shoud be a >= b
fun lcm(a: Long, b: Long): Long = 
	a / gcd(a, b) * b

fun sumDigits(num_: Long): Long {
	var num = num_
	var rtn: Long = 0
	while(num != 0.toLong()) {
		val tmp = num % 10
		rtn += tmp
		num /= 10
	}
	return rtn
}

fun Double.isDecimal(): Boolean = ((this - Math.floor(this)) != 0.0)

// Data Structures

class UnionFind(private val n: Int) {
	private val parent: Array<Int> = Array(n) { 0 }
	private val rank: Array<Int> = Array(n) { 0 }
	private val size: Array<Int> = Array(n) { 0 }

	init {
		for(i in 0 until n) {
			parent[i] = i
			rank[i] = 1
			size[i] = 1
		}
	}

	fun union(u: Int, v: Int) {
		var x = root(u)
		var y = root(v)
		if(x != y) {
			if(rank[y] < rank[x]) {
				val tmp = y 
				y = x
				x = tmp 
			}
			parent[x] = y
			rank[y] += rank[x]
			rank[x] = -1
			val totalSize = size[x] + size[y]
			size[x] = totalSize
			size[y] = totalSize
		}
	}

	fun root(u: Int): Int {
		var x = u
		while(parent[x] != x) {
			x = parent[x]
		}
		return x 
	}

	fun size(u: Int): Int = size[root(u)]
}