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.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;

class Main{
    public static void main(String[] args){
        
        Scanner sc = new Scanner(System.in);
        
        int N = sc.nextInt();
        int M = sc.nextInt();
        int[] a = new int[M];
        int[] b = new int[M];
        int[] c = new int[M];
        int[][] d = new int[N + 1][N + 1];

        for(int i=1; i<=N; i++){
            for(int j=1; j<=N; j++){
                if(i==j){
                    d[i][j] = 0;
                }
                else{
                    d[i][j] = Integer.MAX_VALUE / 2;
                }
            }
        }
        
        for(int i=0; i<M; i++){
            a[i] = sc.nextInt();
            b[i] = sc.nextInt();
            c[i] = sc.nextInt();
            d[a[i]][b[i]] = c[i];
            d[b[i]][a[i]] = c[i];
        }
        int ans = 0;
        for(int k=1; k<=N; k++){
            for(int i=1; i<=N; i++){
                for(int j=1; j<=N; j++){
                    if(d[i][j]>d[i][k]+d[k][j]){
                        d[i][j] = d[i][k] + d[k][j];
                    }
                }
            }
        }
        boolean[] memo = new boolean[M];
        for(int i=1; i<=N; i++){
            for(int j=1; j<=N; j++){
                for(int k=0; k<M; k++){
                    if(d[i][j]==d[i][a[k]]+c[k]+d[b[k]][j] && !memo[k]){
                        ans++;
                        memo[k] = true;
                    }
                }
            }
        }
        System.out.println(M - ans);
    }
}