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.*;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.solve();
    }
    public void solve() {
        int MAX_INF = 10000000;
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int N = scan.nextInt();
        int M = scan.nextInt();
        int[][] cost = new int[N][N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(cost[i], MAX_INF);
        }
        int[] a = new int[M];
        int[] b = new int[M];
        int[] c = new int[M];
        for (int i = 0 ; i < M; i++) {
            a[i] = scan.nextInt()-1;
            b[i] = scan.nextInt()-1;
            c[i] = scan.nextInt();
            cost[a[i]][b[i]] = c[i];
            cost[b[i]][a[i]] = c[i];
        }
        int[][] dp = new int[N][N];
        for(int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                dp[i][j] = MAX_INF;
            }
        }
        for(int i = 0; i < N; i++) {
            for(int j = 0; j < N; j++) {
                if (i == j) {
                    dp[i][j] = 0;
                    continue;
                }
                dp[i][j] = cost[i][j];
            }
        }
        for (int k = 0; k < N; k++) {
            for(int i = 0; i < N; i++) {
                for (int j = 0; j < N; j++) {
                    dp[i][j] = Math.min(dp[i][j],dp[i][k]+dp[k][j]);
                }
            }
        }
        int answer = M;
        for(int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                if (dp[j][a[i]]+c[i] == dp[j][b[i]]) {
                    answer -= 1;
                    break;
                }
            }
        }
        System.out.println(answer);
    }
}