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: Accepted)

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int[][] e=new int[m][3];
        long[][][] warshall=new long[n+1][n][n];
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                warshall[0][i][j]=Integer.MAX_VALUE;
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<3;j++){
                e[i][j]=sc.nextInt();
            }
            warshall[0][e[i][0]-1][e[i][1]-1]=warshall[0][e[i][1]-1][e[i][0]-1]=e[i][2];
        }
        for(int i=1;i<n+1;i++){
            for(int j=0;j<n;j++){
                for(int k=0;k<n;k++){
                    warshall[i][j][k]=Math.min(warshall[i-1][j][k],warshall[i-1][j][i-1]+warshall[i-1][i-1][k]);
                }
            }
        }
        int ans=0;
        for(int i=0;i<m;i++){
            if(warshall[n][e[i][0]-1][e[i][1]-1]<e[i][2]){
                ans++;
            }
        }
        System.out.println(ans);
    }

    
}