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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.*;

class Main {

    void solve() {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), m = in.nextInt();
		int[] a = new int[m], b = new int[m], c = new int[n];
		for(int i = 0; i < m; i++){
			a[i] = in.nextInt() - 1;
			b[i] = in.nextInt() - 1;
			c[i] = in.nextInt();
		}
		int[][] dist = new int[n][n];
		for(int i = 0; i < n; i++){
			Arrays.fill(dist[i], Integer.MAX_VALUE / 3);
			dist[i][i] = 0;
		}
		for(int i = 0; i < m; i++){
			dist[a[i]][b[i]] = Math.min(dist[a[i]][b[i]], c[i]);
			dist[b[i]][a[i]] = Math.min(dist[b[i]][a[i]], c[i]);
		}
		for(int k = 0; k < n; k++){
			for(int i = 0; i < n; i++){
				for(int j = 0; j < n; j++){
					dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
				}
			}
		}
		int ans = m;
		for(int k = 0; k < m; k++){
			boolean f = false;
			for(int i = 0; i < n; i++){
				if(dist[i][a[k]] + c[k] == dist[i][b[k]]){
					f = true;
				}
			}
			if(f){
				ans--;
			}
		}
		System.out.println(ans);
	}

    public static void main(String[] args) {
        new Main().solve();
    }
}