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.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(),m = sc.nextInt();
        int inf = Integer.MAX_VALUE/2-10;
        int[][] dist = new int[n+1][n+1];
        for(int i=0;i<=n;i++) Arrays.fill(dist[i],inf);
        for(int i=0;i<=n;i++) dist[i][i] = 0;
        Edge[] dic = new Edge[m];
        for(int i=0;i<m;i++){
            int a = sc.nextInt(), b=sc.nextInt(), c = sc.nextInt();
            dic[i] = new Edge(a,b,c);
            dist[a][b] = Math.min(dist[a][b],c);
            dist[b][a] = Math.min(dist[b][a],c);
        }
        // run Floyd-Warshall
        for(int mid=1;mid<=n;mid++){
            for(int i=1;i<=n;i++){
                for(int j=1;j<=n;j++) dist[i][j] = Math.min(dist[i][j], dist[i][mid]+dist[mid][j]);
            }
        }
        int ans = 0;
        for(Edge e:dic){
            if(dist[e.left][e.right]<e.length) ans++;
        }
        System.out.println(ans);
    }
    static class Edge{
        int left,right,length;
        public Edge(int l, int r, int len){
            left = l;
            right = r;
            length = len;
        }
    }
}