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.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class Main {
    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final int m = scanner.nextInt();

        final int[][] graph = new int[n + 1][n + 1];
        for (final int[] array : graph) {
            Arrays.fill(array, Integer.MAX_VALUE / 2);
        }
        final List<Edge> edges = new LinkedList<>();
        for (int i = 0; i < m; i++) {
            final int from = scanner.nextInt();
            final int to = scanner.nextInt();
            final int cost = scanner.nextInt();

            edges.add(new Edge(from, to, cost));
            graph[from][to] = cost;
            graph[to][from] = cost;
        }

        for (int k = 0; k < n + 1; k++) {
            for (int i = 0; i < n + 1; i++) {
                for (int j = 0; j < n + 1; j++) {
                    graph[i][j] = Math.min(graph[i][j], graph[i][k] + graph[k][j]);
                }
            }
        }

        final long answer = edges.stream()
                .filter(edge -> edge.cost > graph[edge.from][edge.to])
                .count();

        System.out.println(answer);
    }

    static class Edge {
        int from;
        int to;
        int cost;

        Edge(final int from, final int to, final int cost) {
            this.from = from;
            this.to = to;
            this.cost = cost;
        }
    }
}