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.util.*;
import java.io.*;
class Main {
    static HashSet<String> hs;
    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());

        int N = Integer.parseInt(st.nextToken());
        int M = Integer.parseInt(st.nextToken());
        HashMap<Integer, List<Edge>> hm = new HashMap();
        hs = new HashSet();
        for (int i=0; i<M; i++) {
            st = new StringTokenizer(br.readLine());
            int start = Integer.parseInt(st.nextToken())-1;
            int end = Integer.parseInt(st.nextToken())-1;
            int wt = Integer.parseInt(st.nextToken());
            List<Edge> lst = hm.getOrDefault(start, new LinkedList());
            lst.add(new Edge(start, end,wt));
            hm.put(start, lst);
            lst = hm.getOrDefault(end, new LinkedList());
            lst.add(new Edge(end, start,wt));
            hm.put(end, lst);
        }
        for(int i=0; i<N; i++){
            dijkstra(hm, i, N);
        }
        System.out.println(M-hs.size());

    }
    public static class Edge{
        int src,dest,weight;
        Edge(int s, int x, int y){
            src=s;
            dest=x;
            weight=y;
        }
    }

    private static void dijkstra(HashMap<Integer, List<Edge>> graph, 
                                        int startVertex, int n) 
    { 
        int[] shortestDist = new int[n];

        PriorityQueue<Edge> pq = new PriorityQueue<>(new Comparator<Edge>(){
            public int compare(Edge a, Edge b){
                return a.weight-b.weight;
            }
        });

        for(int i=0; i<n; i++)
            shortestDist[i]=Integer.MAX_VALUE;

        shortestDist[startVertex]=0;
        HashSet<Integer> visited = new HashSet();
        
        int curVertex = startVertex;

        for(Edge e: graph.get(curVertex)){
            pq.offer(e);
        }
        visited.add(curVertex);
        while(!pq.isEmpty()){
            
            Edge curEdge = pq.poll();
            curVertex = curEdge.src;
            if(visited.contains(curEdge.dest))
                continue;

            visited.add(curEdge.dest);
            if(shortestDist[curEdge.dest]>shortestDist[curVertex]+curEdge.weight){
                shortestDist[curEdge.dest]=shortestDist[curVertex]+curEdge.weight;
                hs.add(Math.min(curEdge.src,curEdge.dest)+","+Math.max(curEdge.src,curEdge.dest));
                for(Edge e: graph.get(curEdge.dest)){
                    pq.offer(new Edge(e.src, e.dest, e.weight+shortestDist[e.src]));
                }
            }

        }

    }
}