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: Wrong Answer)


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt(); //頂点
		int m = sc.nextInt(); //辺
		int[][] dist = new int[n+1][n+1];
		int[] a = new int[m];
		int[] b = new int[m];
		int[][] edge = new int[n+1][n+1];
		boolean[][] count=new boolean[n+1][n+1];
		int ans=0;
		
		for (int i = 0; i < m; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
			edge[b[i]][a[i]]=edge[a[i]][b[i]] = sc.nextInt();
		}
		for (int i = 1; i < n+1; i++) {
			for (int j = 1; j < n+1; j++) {
				dist[i][j] = edge[i][j];
			}
		}
		// ワーシャルフロイド法で最短距離を計算
		for (int k = 1; k < n+1; k++) {
			for (int i = 1; i < n+1; i++) {
				for (int j = 1; j < n+1; j++) {
						dist[i][j] = Math.min(dist[i][j],dist[i][k] + dist[k][j]);
				}
			}
		}
		for (int s = 1; s < n+1; s++) {
			for (int t = 1; t < n+1; t++) {
				for (int i = 0; i < m; i++) {
					if (dist[s][t] == dist[s][a[i]] + edge[a[i]][b[i]] + dist[b[i]][t] ||
							dist[s][t] == dist[s][b[i]] + edge[b[i]][a[i]] + dist[a[i]][t] ) {
						count[a[i]][b[i]]=true;
					}
				}
			}
		}
		for(int i=0;i<m;i++){
			if(count[a[i]][b[i]] || count[b[i]][a[i]])ans++;
		}
		System.out.println(m-ans);
	}
}