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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

namespace ABC051
{

    class Node
    {
        public int NextNode { get; }
        public int Cost { get; }

        public Node(int next, int cost)
        {
            NextNode = next;
            Cost = cost;
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            string[] input;
            input = Console.ReadLine().Split(' ');
            int N = int.Parse(input[0]);
            int M = int.Parse(input[1]);

            var nodelist = Enumerable.Range(0, N).Select(x => new List<Node>()).ToArray();
            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 cost = int.Parse(input[2]);

                nodelist[a].Add(new Node(b, cost));
                //nodelist[b].Add(new Node(a, cost));
            }

            int[] costlist = new int[N].Select(x => int.MaxValue).ToArray();
            costlist[0] = 0;
            var que = new Queue<int>();
            que.Enqueue(0);

            var nl = new List<Node>();
            while (que.Any())
            {
                int p = que.Dequeue();

                foreach (var next in nodelist[p])
                {
                    //Console.WriteLine("costlist[next.NextNode] : " + costlist[next.NextNode]);
                    //Console.WriteLine(" next.Cost : " + next.Cost);
                    //Console.WriteLine("costlist[p] : " + costlist[p]);

                    if (costlist[next.NextNode] > next.Cost + costlist[p])
                    {
                        costlist[next.NextNode] = costlist[p] + next.Cost;
                        que.Enqueue(next.NextNode);
                        nl.Add(next);
                        //Console.WriteLine("Count : " + nl.Count());
                    }
                }
                foreach (Node n in nl)
                {
                    //Console.WriteLine(nodelist[p].Count());
                    nodelist[p].Remove(n);
                    //Console.WriteLine(nodelist[p].Count());
                }
            }

            int result = M - nl.Count();
            //foreach (List<Node> n in nodelist)
            //{
            //    result = n.Count;
            //}
            Console.WriteLine(result);
        }
    }
}