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)

#region using
using System;
using System.Collections.Generic;
using System.Linq;
using IEnumerable = System.Collections.IEnumerable;
using IEnumerator = System.Collections.IEnumerator;
using BitArray = System.Collections.BitArray;
using BigInteger = System.Numerics.BigInteger;
using TextReader = System.IO.TextReader;
using System.Text;
#endregion

namespace AtCoderProject
{
    public class Program
    {
        public object Calc()
        {
            var N = consoleReader.Int;
            var M = consoleReader.Int;
            graph = new LengthGraphBuilder(N, consoleReader, M).ToArray();
            lengthes = new int[N, N];
            for (int i = 0; i < N; i++)
                for (int j = 0; j < N; j++)
                    lengthes[i, j] = 10000000;
            for (int i = 0; i < N; i++)
                UpdateLength(i, i, 0);
            var count = 0;
            for (int i = 0; i < N; i++)
                foreach (var next in graph[i])
                {
                    if (lengthes[i, next.to] < next.length) count++;
                }
            return count / 2;
        }
        Next[][] graph;
        int[,] lengthes;

        void UpdateLength(int from, int to, int length)
        {
            if (lengthes[from, to] < length) return;
            lengthes[from, to] = lengthes[to, from] = length;
            foreach (var next in graph[to])
                UpdateLength(from, next.to, length + next.length);
        }


        struct Next
        {
            public int to;
            public int length;
            public override string ToString() => $"to: {to} length:{length}";
        }
        class LengthGraphBuilder
        {
            public List<Next>[] lists;
            public LengthGraphBuilder(int count)
            {
                this.lists = new List<Next>[count];
                for (int i = 0; i < count; i++)
                    this.lists[i] = new List<Next>();
            }
            public LengthGraphBuilder(int count, ConsoleReader cr, int edgeCount) : this(count)
            {
                for (int i = 0; i < edgeCount; i++)
                    this.Add(cr.Int - 1, cr.Int - 1, cr.Int);
            }
            public void Add(int from, int to, int length)
            {
                this.lists[from].Add(new Next { to = to, length = length });
                this.lists[to].Add(new Next { to = from, length = length }); // 有向の場合は消す
            }
            public Next[][] ToArray()
            {
                var N = this.lists.Length;
                var res = new Next[N][];
                for (int i = 0; i < N; i++)
                    res[i] = this.lists[i].ToArray();
                return res;
            }
        }


        #region いつもの
#pragma warning disable
        [System.Diagnostics.DebuggerBrowsable(System.Diagnostics.DebuggerBrowsableState.Never)] private ConsoleReader consoleReader;
        public Program(ConsoleReader consoleReader) { this.consoleReader = consoleReader; }
        static void Main() => Console.WriteLine(new Program(new ConsoleReader(Console.In)).Calc()); static string AllLines<T>(IEnumerable<T> source) => string.Join("\n", source);
    }
    static class Ext
    {
        public static Dictionary<TKey, int> GroupCount<TSource, TKey>(this IEnumerable<TSource> source, Func<TSource, TKey> keySelector) => source.GroupBy(keySelector).ToDictionary(g => g.Key, g => g.Count());
        public static Dictionary<TKey, int> GroupCount<TKey>(this IEnumerable<TKey> source) => source.GroupCount(i => i);
    }
    public class ConsoleReader { private string[] ReadLineSplit() => textReader.ReadLine().Split(Array.Empty<char>(), StringSplitOptions.RemoveEmptyEntries); private string[] line = Array.Empty<string>(); private int linePosition; private TextReader textReader; public ConsoleReader(TextReader tr) { textReader = tr; } public int Int => int.Parse(String); public long Long => long.Parse(String); public double Double => double.Parse(String); public string String { get { if (linePosition >= line.Length) { linePosition = 0; line = ReadLineSplit(); } return line[linePosition++]; } } public class SplitLine { private string[] splited; public SplitLine(ConsoleReader cr) { splited = cr.ReadLineSplit(); cr.line = Array.Empty<string>(); } public int[] Int => String.Select(x => int.Parse(x)).ToArray(); public int[] Int0 => String.Select(x => int.Parse(x) - 1).ToArray(); public long[] Long => String.Select(x => long.Parse(x)).ToArray(); public double[] Double => String.Select(x => double.Parse(x)).ToArray(); public string[] String => splited; } public SplitLine Split => new SplitLine(this); public class RepeatReader : IEnumerable<ConsoleReader> { ConsoleReader cr; int count; public RepeatReader(ConsoleReader cr, int count) { this.cr = cr; this.count = count; } public IEnumerator<ConsoleReader> GetEnumerator() => Enumerable.Repeat(cr, count).GetEnumerator(); System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator() => GetEnumerator(); public IEnumerable<string> String => this.Select(cr => cr.String); public IEnumerable<int> Int => this.Select(cr => cr.Int); public IEnumerable<int> Int0 => this.Select(cr => cr.Int - 1); public IEnumerable<long> Long => this.Select(cr => cr.Long); public IEnumerable<double> Double => this.Select(cr => cr.Double); } public RepeatReader Repeat(int count) => new RepeatReader(this, count); }
    #endregion
}