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 = Integer.parseInt(sc.next());
        int M = Integer.parseInt(sc.next());
        int[] ai = new int[M];
        int[] bi = new int[M];
        int[] ci = new int[M];
        for (int i = 0; i < M; i++) {

            ai[i] = Integer.parseInt(sc.next());
            bi[i] = Integer.parseInt(sc.next());
            ci[i] = Integer.parseInt(sc.next());

        }
        sc.close();

        int max = 100000;

        //最大の距離は1000*99なのでintで足りるはず
        int[][] memo      = new int[N][N];//最短距離を記録するメモ

        //3重のfor文を用いて2点間の最小の値を探していくダイクストラ法
        for (int i = 0; i < N; i++) {

            for (int j = 0; j < N; j++) {

                if (i == j) {

                    memo[i][j] = 0;
                } else {

                    memo[i][j] = max;
                }
            }
        }
        //既知の経路を代入
        for (int i = 0; i < M; i++) {

            memo[ai[i]-1][bi[i]-1] = ci[i];
            memo[bi[i]-1][ai[i]-1] = ci[i];
        }

        //頂点1,2,・・N-1と頂点1+1,1+2,・・N,・2+1,2+2・・・N,・・・を結ぶ全パターンをメモ
        for (int i = 0; i < N; i++) {

            for (int j = 0; j < N; j++) {

                for (int k = 0; k < N; k++) {

                    if (memo[i][j] > memo[i][k] + memo[j][k]) {

                        memo[i][j] = memo[i][k] + memo[j][k];
                        memo[j][i] = memo[i][k] + memo[j][k];
                    }

                }
            }
        }

        //使われなかった経路の本数
        int ans = 0;

        for (int i = 0; i < M; i++) {

            if (ci[i] > memo[ai[i]-1][bi[i]-1]) {

                ans++;
            }
        }

        System.out.println(ans);

    }

}