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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.io.*;
import java.util.*;
import static java.lang.System.out;

public class Main {
    static MyReader in = new MyReader();

    public static void main(String[] args) {
        Graph g = new Graph(in);
        int ans = g.solve();
        out.println(ans);
    }
}

class Graph {
    private static final int INF = Integer.MAX_VALUE;

    int V;
    int E;

    ArrayList<int[]>[] T;
    int[][] d;
    IntQueue[][] q;
    boolean[][] unused;

    Graph(MyReader in) {
        V = in.i();
        E = in.i();
        T = new ArrayList[V];
        d = new int[V][V];
        q = new IntQueue[V][V];
        unused = new boolean[V][V];

        for (int i = 0; i < V; i++) {
            q[i][i] = new IntQueue(V);
            for (int j = i + 1; j < V; j++) {
                (q[i][j] = new IntQueue(V)).push(i);
                q[i][j].push(j);
                (q[j][i] = new IntQueue(V)).push(j);
                q[j][i].push(i);
                d[i][j] = d[j][i] = INF;
            }
        }

        Arrays.setAll(T, i -> new ArrayList<>());

        for (int i = 0; i < E; i++) {
            int u = in.i() - 1;
            int v = in.i() - 1;
            int w = in.i();
            d[u][v] = d[v][u] = w;
            T[u].add(new int[]{v, w});
            T[v].add(new int[]{u, w});
            unused[u][v] = unused[v][u] = true;
        }
    }

    int solve() {
        WarshallFloyd();
        for (int i = 0; i < V; i++) {
            for (int j = i + 1; j < V; j++) {
                int u = q[i][j].pop();
                IntQueue p = q[i][j];
                while (!p.isEmpty()) {
                    int v = p.pop();
                    unused[u][v] = unused[v][u] = false;
                    u = v;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < V; i++) {
            for (int j = i + 1; j < V; j++) {
                if (unused[i][j]) {
                    ans++;
                }
            }
        }
        return ans;
    }

    int[] dijkstra(int s) {
        int[] d = new int[V];
        Arrays.fill(d, INF);
        d[s] = 0;

        PriorityQueue<Integer> A = new PriorityQueue<>((x, y) -> d[x] - d[y]);
        A.add(s);

        while (!A.isEmpty()) {
            int v = A.poll();
            for(int[] edge : T[v]){
                int u = edge[0];
                int w = edge[1];
                if(d[u] == INF){
                    d[u] = d[v] + w;
                    A.add(u);
                } else if(d[u] - d[v] > w){
                    d[u] = d[v] + w;
                }
            }
        }
        return d;
    }

    int[][] WarshallFloyd() {
        for (int k = 0; k < V; k++) {
            for (int i = 0; i < V; i++) {
                for (int j = 0; j < V; j++) {
                    int t = d[i][k] + d[k][j];
                    if (t < d[i][j]) {
                        d[i][j] = t;
                        q[i][j].clear();
                        q[i][j].pushAll(q[i][k]);
                        q[i][j].unpush();
                        q[i][j].pushAll(q[k][j]);

                    }
                }
            }
        }
        return d;
    }
}

class IntQueue {
    final int MAX;
    int[] array;
    int head = 0;
    int tail = 0;
    int count = 0;

    IntQueue(int N) {
        MAX = N;
        array = new int[N];
    }

    void push(int n) {
        array[tail] = n;
        if (++tail == MAX) tail = 0;
        count++;
    }

    void unpush() {
        if (--tail == -1) tail += MAX;
        count--;
    }

    void pushAll(IntQueue q) {
        int n = q.size();
        for (int i = 0; i < n; i++) {
            push(q.array[(i + head) % MAX]);
        }
    }

    int pop() {
        int n = array[head];
        if (++head == MAX) head = 0;
        count--;
        return n;
    }

    void clear() {
        head = tail = count = 0;
    }

    boolean isEmpty() {
        return count == 0;
    }

    int size() {
        return count;
    }

    public String toString() {
        return Arrays.toString(array);
    }
}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = super.read(cbuf, 0, cbuf.length);
            } catch (IOException e) {
                e.printStackTrace();
            }
            head = 0;
        }
        return cbuf[head++];
    }

    void back() {
        head--;
    }

    boolean minus() {
        boolean minus;
        while (true) {
            char c = next();
            if (!isDelimiter(c)) {
                if (!(minus = c == '-')) back();
                return minus;
            }
        }
    }

    void skip() {
        while (isDelimiter(next()));
        back();
    }

    char[] s(int N) {
        char[] cbuf = new char[N];
        read(cbuf, 0, N);
        return cbuf;
    }

    public int read(char[] cbuf, int off, int len) {
        skip();
        int i;
        for (i = 0; i < cbuf.length; i++) {
            char c = next();
            if (isDelimiter(c)) {
                break;
            }
            cbuf[i] = c;
        }
        return i;
    }

    boolean isDelimiter(char c) {
        return c == ' ' || c == '\n' || c == '\r';
    }

    int i() {
        boolean minus = minus();
        int n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        boolean minus = minus();
        long n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }
}