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 Go (1.6) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

const (
	initialBufSize = 100000
	maxBufSize     = 10000000
)

var (
	sc = bufio.NewScanner(os.Stdin)
	wt = bufio.NewWriter(os.Stdout)
)

func next() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	i, _ := strconv.Atoi(next())
	return i
}

func puts(a ...interface{}) {
	fmt.Fprintln(wt, a...)
}

func min(nums ...int) int {
	ret := nums[0]
	for _, v := range nums {
		if v < ret {
			ret = v
		}
	}
	return ret
}

func main() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, initialBufSize), maxBufSize)
	defer wt.Flush()

	const inf = 1 << 60

	n, m := nextInt(), nextInt()

	cost, prev, use := make([][]int, n), make([][]int, n), make([][]bool, n)
	for i := 0; i < n; i++ {
		cost[i], prev[i], use[i] = make([]int, n), make([]int, n), make([]bool, n)
		for j := 0; j < n; j++ {
			cost[i][j], prev[i][j] = inf, i
		}
	}
	for i := 0; i < m; i++ {
		a, b, c := nextInt()-1, nextInt()-1, nextInt()
		cost[a][b], cost[b][a] = c, c
	}

	// ワーシャルフロイド
	for k := 0; k < n; k++ {
		for i := 0; i < n; i++ {
			for j := 0; j < n; j++ {
				if cost[i][k]+cost[k][j] < cost[i][j] {
					cost[i][j] = cost[i][k] + cost[k][j]
					prev[i][j] = prev[k][j]
				}
			}
		}
	}

	for start := 0; start < n; start++ {
		for goal := start + 1; goal < n; goal++ {
			for cur := goal; cur != start; cur = prev[start][cur] {
				use[prev[start][cur]][cur], use[cur][prev[start][cur]] = true, true
			}
		}
	}

	useCnt := 0
	for i := range use {
		for j := range use[i] {
			if use[i][j] {
				useCnt++
			}
		}
	}

	puts(m - useCnt/2)
}