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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.List;

public class Main {
	static List<Integer> path;

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int m = Integer.parseInt(sa[1]);
		int[] a = new int[m];
		int[] b = new int[m];
		int[] c = new int[m];
		int[][] ab = new int[n][n];
		for (int i = 0; i < n; i++) {
			Arrays.fill(ab[i], 100000000);
		}
		for (int i = 0; i < m; i++) {
			sa = br.readLine().split(" ");
			a[i] = Integer.parseInt(sa[0]) - 1;
			b[i] = Integer.parseInt(sa[1]) - 1;
			c[i] = Integer.parseInt(sa[2]);
			ab[a[i]][b[i]] = c[i];
			ab[b[i]][a[i]] = c[i];
		}
		br.close();

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

		int ans = 0;
		for (int k = 0; k < m; k++) {
			boolean flg = false;
			hen:
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < n; j++) {
					int ak = a[k];
					int bk = b[k];
					if (ab[i][j] == ab[i][ak] + c[k] + ab[bk][j]) {
						flg = true;
						break hen;
					}
				}
			}
			if (!flg) {
				ans++;
			}
		}
		System.out.println(ans);
	}
}