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 (
	"fmt"
)

func main() {
	var N, M int
	fmt.Scan(&N, &M)
	a, b, c := make([]int, M), make([]int, M), make([]int, M)
	for i := 0; i < M; i++ {
		fmt.Scan(&a[i], &b[i], &c[i])
		a[i]--
		b[i]--
	}

	dist := make([][]int, N)
	for i := 0; i < N; i++ {
		dist[i] = make([]int, N)
	}

	for i := 0; i < N; i++ {
		for j := 0; j < N; j++ {
			if i == j {
				continue
			}

			dist[i][j] = 1e9
		}
	}

	for i := 0; i < M; i++ {
		dist[a[i]][b[i]] = c[i]
		dist[b[i]][a[i]] = c[i]
	}

	for k := 0; k < N; k++ {
		for i := 0; i < N; i++ {
			for j := 0; j < N; j++ {
				if dist[i][k] + dist[k][j] < dist[i][j] {
					dist[i][j] = dist[i][k] + dist[k][j]
				}
			}
		}
	}

	count := 0
	for k := 0; k < M; k++ {
		flag := true
		for i := 0; i < N; i++ {
			for j := 0; j < N; j++ {
				if i == j {
					continue
				}

				if dist[i][a[k]] + c[k] + dist[b[k]][j] == dist[i][j] {
					flag = false
				}
			}
		}

		if flag {
			count++
		}
	}

	fmt.Println(count)
}