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 {
	static class Edge {
		int src;
		int dst;
		int cost;

		public Edge(int src, int dst, int cost) {
			this.src = src;
			this.dst = dst;
			this.cost = cost;
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		ArrayList<Edge> es = new ArrayList<>();

		int[][] d = new int[N][N];
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j < N; ++j) {
				if (i == j)
					continue;
				d[i][j] = Integer.MAX_VALUE / 10;
			}
		}
		for (int i = 0; i < M; ++i) {
			int a = sc.nextInt() - 1;
			int b = sc.nextInt() - 1;
			int c = sc.nextInt();
			d[a][b] = c;
			d[b][a] = c;
			es.add(new Edge(a, b, c));
			es.add(new Edge(b, a, c));
		}
		for (int m = 0; m < N; ++m) {
			for (int i = 0; i < N; ++i) {
				for (int j = 0; j < N; ++j) {
					d[i][j] = Math.min(d[i][m] + d[m][j], d[i][j]);
				}
			}
		}
		int ans = 0;
		out: for (Edge e : es) {
			for (int i = 0; i < N; ++i) {
				for (int j = 0; j < N; ++j) {
					if (d[i][e.src] + e.cost + d[e.dst][j] == d[i][j])
						continue out;
				}
			}
			++ans;
		}
		System.out.println(ans / 2);
	}
}