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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;

class program
{



    static void Main(string[] args)
    {
        //dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j])

        string[] input = Console.ReadLine().Split();
        int N = int.Parse(input[0]);
        int M = int.Parse(input[1]);

        int[,] graph = new int[N, N];
        int[,] dp = new int[N, N];

        int ans = 0;

        for(int i = 0; i < N; i++)
        {
            for(int j = 0; j < N; j++)
            {
                dp[i, j] = 100000000;
                graph[i, j] = -1;
            }
        }

        for(int i = 0; i < M; i++)
        {
            input = Console.ReadLine().Split();
            int a = int.Parse(input[0]) - 1;
            int b = int.Parse(input[1]) - 1;
            int c = int.Parse(input[2]);
            graph[a, b] = c;
            dp[a, b] = c;
            dp[b, a] = c;
        }

        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]);
                }
            }
        }

        for(int i = 0; i < N; i++)
        {
            for(int j = 0; j < N; j++)
            {
                if (graph[i, j] > 0)
                {
                    if (graph[i, j] > dp[i, j]) ans++;
                }
            }
        }

        Console.WriteLine(ans);

    }
}