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.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.LinkedList;
import java.util.List;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;
import java.util.ArrayDeque;
import java.util.Queue;
class Edge{
    int to;
    long cost;
    Edge(int to,long cost){
        this.to = to;
        this.cost = cost;
    }
}

public class Main {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int n = fs.nextInt();
        int m = fs.nextInt();
      	int count = 0;
        int[] a = new int[n];
        int[][] w = new int[n][n];
        boolean[][] ans = new boolean[n][n];
        List<LinkedList<Edge>> graph = new ArrayList<>();
        List<LinkedList<Edge>> select = new ArrayList<>();
        boolean[] visited = new boolean[n];
        boolean[] z = new boolean[n];
        int[] kanban = new int[n];
        for(int i = 0; i < n; i++){
          for(int j = 0; j < n; j++){
            ans[i][j] = false;
          }
        }
        for(int i=0;i<n;i++){
            graph.add(new LinkedList<>());
            select.add(new LinkedList<>());
        }
        int from,to,cost;
        for(int i=0;i<m;i++){
            from = fs.nextInt()-1;
            to = fs.nextInt()-1;
            cost = fs.nextInt();
            graph.get(from).add(new Edge(to,cost));
            graph.get(to).add(new Edge(from,cost));
        }
        for(int i = 0; i < n; i++){
          long inf = 1000000000000L;
          long[] dist = new long[n];
          Arrays.fill(dist,inf);
          Arrays.fill(visited,true);
          Arrays.fill(z,true);
          Queue<Integer> queue = new ArrayDeque<>();
          queue.add(i);
          dist[i] = 0;
          visited[i] = false;
          Arrays.fill(kanban,-1);
        while(!queue.isEmpty()){
            int x = queue.poll();
          	z[x] = false;
            for(Edge e : graph.get(x)){
              if(visited[e.to] == true){
                visited[e.to] = false;
                queue.add(e.to);
              }
              if(Math.min(e.cost+dist[x],dist[e.to]) == e.cost+dist[x]){
                if(kanban[e.to] != -1){
                  w[e.to][kanban[e.to]]++;
                  w[kanban[e.to]][e.to]++;
                }
                kanban[e.to] = x;
              }
              else{
                if(z[e.to] == true){
                  w[e.to][x]++;
                  w[x][e.to]++;
                }
              }
              dist[e.to] = Math.min(e.cost+dist[x],dist[e.to]);
            }  
        }
          for(int q = 0; q < n; q++){
              int j = kanban[q];
              if(j == -1){
                continue;
              }
              if(!ans[q][j]){
                count += 1;
                ans[q][j] = true;
                ans[j][q] = true;
                }
              }
              }
      System.out.println(m - count);
}
}
class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte(){
        if(ptr < buflen){
            return true;
        }else{
            ptr = 0;
            try{
                buflen = in.read(buffer);
            }catch(IOException e){
                e.printStackTrace();
            }
            if(buflen <=0){
                return false;
            }
        }
        return true;
    }
 
    private int readByte(){
        if(hasNextByte())return buffer[ptr++];
        else return -1;
    }
 
    private static boolean isPrintableChar(int c){
        return 33<=c && c<=126;
    }
    public boolean hasNext(){
        while(hasNextByte() && !isPrintableChar(buffer[ptr]))ptr++;
        return hasNextByte();
    }
 
    public String next(){
        if(!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)){
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
 
    public long nextLong(){
        if(!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if(b == '-'){
            minus = true;
            b = readByte();
        }
        if(b < '0' || '9' < b){
            throw new NumberFormatException();
        }
        while(true){
            if('0' <= b && b<='9'){
                n*=10;
                n+=b-'0';
            }else if(b==-1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt(){
        long nl = nextLong();
        if(nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDoutble(){return Double.parseDouble(next());}
}