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.Scanner;
import java.util.*;
public class Main {
    static int INF = 1000000000;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[][] cost = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cost[i][j] = INF;
            }
        }

        int[] a = new int[m];
        int[] b = new int[m];
        int[] c = new int[m];
        for (int i = 0; i < m; i++) {
            a[i] = sc.nextInt() - 1;
            b[i] = sc.nextInt() - 1;
            c[i] = sc.nextInt();
            cost[a[i]][b[i]] = c[i];
            cost[b[i]][a[i]] = c[i];
        }

        int[][] dist = new int[n][n];
        for (int i = 0; i < n; i++) {
            int[] d = dijkstra(i, n, cost);
            for (int j = 0; j < n; j++) {
                dist[i][j] = d[j];
            }
        }
        int ans = solve(n, m, dist, a, b, c);
        System.out.println(ans);
    }
    public static int solve(int n, int m, int[][] dist, int[] a, int b[], int c[]) {
        int ans = m;
        for (int i = 0; i < m; i++) {
            boolean ok = false;
            for (int j = 0; j < n; j++) {
                if (dist[j][a[i]] + c[i] == dist[j][b[i]]) {
                    ok = true;
                }
            }
            if (ok) {
                ans--;
            }
        }
        return ans;
    }
    public static int[] dijkstra(int from, int n, int[][] cost) {
        int[] d = new int[n];
        d[from] = 0;
        for (int i = 0; i < n; i++) {
            if (i == from) {
                continue;
            }
            d[i] = INF;
        }

        boolean[] visited = new boolean[n];
        while (true) {
            int v = -1;
            for (int u = 0; u < n; u++) {
                if (!visited[u] && (v == -1 || d[u] < d[v])) {
                    v = u;
                }
            }
            if (v == -1) {
                break;
            }
            visited[v] = true;

            for (int u = 0; u < n; u++) {
                d[u] = Math.min(d[u], d[v] + cost[v][u]);
            }
        }
        return d;
    }
}