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

/**
 * Created by ryosuke on 2017/04/21.
 */
public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int m = scanner.nextInt();
        int[][] edge = new int[m][3];
        int[][] cost = new int[n][n];
        ArrayList<Integer>[][] minEdgeList = new ArrayList[n][n];

        for (int i = 0; i < m; i++) {
            edge[i][0] = scanner.nextInt()-1;
            edge[i][1] = scanner.nextInt()-1;
            edge[i][2] = scanner.nextInt();
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                minEdgeList[i][j] = new ArrayList<>();
            }
            Arrays.fill(cost[i], Integer.MAX_VALUE);
        }

        for (int i = 0; i < n; i++) {
            cost[i][i] = 0;
        }

        // ベルマンフォード法？
        for (int startIndex = 0; startIndex < n; startIndex++) {
            while (true) {
                boolean isUpdated = false;
                for (int edgeIndex = 0; edgeIndex < m; edgeIndex++) {
                    int viaNode = edge[edgeIndex][0];
                    int targetNode = edge[edgeIndex][1];
                    int viaCost = edge[edgeIndex][2];
                    if (cost[startIndex][targetNode] > cost[startIndex][viaNode] + viaCost) {
                        isUpdated = true;
                        cost[startIndex][targetNode] = cost[startIndex][viaNode] + viaCost;
                        minEdgeList[startIndex][targetNode].clear();
                        minEdgeList[startIndex][targetNode].addAll(minEdgeList[startIndex][viaNode]);
                        minEdgeList[startIndex][targetNode].add(edgeIndex);
                    }
                    viaNode = edge[edgeIndex][1];
                    targetNode = edge[edgeIndex][0];
                    if (cost[startIndex][targetNode] > cost[startIndex][viaNode] + viaCost) {
                        isUpdated = true;
                        cost[startIndex][targetNode] = cost[startIndex][viaNode] + viaCost;
                        minEdgeList[startIndex][targetNode].clear();
                        minEdgeList[startIndex][targetNode].addAll(minEdgeList[startIndex][viaNode]);
                        minEdgeList[startIndex][targetNode].add(edgeIndex);
                    }
                }
                if(!isUpdated) {
                    break;
                }
            }
        }

        boolean[] used = new boolean[m];
        Arrays.fill(used, false);

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                for (int edgeIndex : minEdgeList[i][j]) {
                    used[edgeIndex] = true;
                }
            }
        }

        int ans = 0;
        for (boolean isUsed : used) {
            if (!isUsed) ans++;
        }
        System.out.println(ans);
    }
}