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.*;

// ABC 51-D
// https://abc051.contest.atcoder.jp/tasks/abc051_d

public class Main {

	static int N;
	static int[][] map;
	static int[][] usedEdges;
	static int numUsedEdges;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		
		N = in.nextInt();
		int M = in.nextInt();
		map = new int[N][N];
		usedEdges = new int[N][N];
		numUsedEdges = 0;
		
		for (int i = 0; i < M; i++) {
			int a = in.nextInt() - 1;
			int b = in.nextInt() - 1;
			int c = in.nextInt();
			
			map[a][b] = c;
			map[b][a] = c;
		}
		
		for (int i = 0; i < N; i++) {
			for (int j = i + 1; j < N; j++) {
				dijkstra(i, j);
			}
		}
		System.out.println(M - numUsedEdges);
	}
	
	public static void dijkstra(int start, int end) {
		int[] visited = new int[N];
		PriorityQueue<Custom_Pair> queue = new PriorityQueue<Custom_Pair>();
		queue.add(new Custom_Pair(start, 0));
		
		while (queue.size() > 0) {
			Custom_Pair p = queue.poll();
			
			if (visited[p.p] == 1) {
				continue;
			}
			visited[p.p] = 1;
			p.path.add(p.p);
			
			if (p.p == end) {
//				System.out.printf("(%d, %d)\n", start, end);
				for (int i = 0; i < p.path.size() - 1; i++) {
					int a = p.path.get(i);
					int b = p.path.get(i + 1);
//					System.out.printf("  %d => %d", a, b);
					if (usedEdges[a][b] == 0 && usedEdges[b][a] == 0) {
						usedEdges[a][b] = 1;
						usedEdges[b][a] = 1;
						numUsedEdges++;
//						System.out.printf(" (NEW)");
					}
//					System.out.println();
				}
				break;
			}
			
			for (int i = 0; i < N; i++) {
				if (map[p.p][i] > 0 && visited[i] == 0) {
					try {
						Custom_Pair newPair = (Custom_Pair) p.clone();
						newPair.p = i;
						newPair.cost += map[p.p][i];
						queue.add(newPair);
					} catch (Exception e) {
						
					}
				}
			}
		}
	}
}

class Custom_Pair implements Comparable<Custom_Pair> {
	int p;
	int cost;
	ArrayList<Integer> path;
	
	public Custom_Pair(int p, int cost) {
		super();
		this.p = p;
		this.cost = cost;
		this.path = new ArrayList<Integer>();
	}

	@Override
	public int compareTo(Custom_Pair o) {
		return this.cost - o.cost;
	}

	@Override
	protected Object clone() throws CloneNotSupportedException {
		Custom_Pair p = new Custom_Pair(this.p, this.cost);
		for (int i = 0; i < this.path.size(); i++) {
			p.path.add(this.path.get(i));
		}
		return p;
	}
}