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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

int main() {
  int N, M; cin >> N >> M;
  const int INF = 1000000007;
  vector<vector<int>> G(N, vector<int>(N, INF));
  for (int i = 0; i < N; i++) G.at(i).at(i) = 0;
  vector<int> A(M), B(M), C(M);
  for (int i = 0; i < M; i++) {
    int a, b, c; cin >> a >> b >> c;
    a--; b--;
    G.at(a).at(b) = c;
    G.at(b).at(a) = c;
    A.at(i) = a;
    B.at(i) = b;
    C.at(i) = c;
  }
  for (int k = 0; k < N; k++) {
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        G.at(i).at(j) = min(G.at(i).at(j), G.at(i).at(k) + G.at(k).at(j));
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < M; i++) {
    bool ok = true;
    for (int s = 0; s < N; s++) {
      for (int t = 0; t < N; t++) {
        if (s == t) continue;
        if (G.at(s).at(t) == G.at(s).at(A.at(i)) + C.at(i) + G.at(B.at(i)).at(t)) {
          ok = false;
        }
      }
    }
    if (ok) ans++;
  }
  cout << ans << endl;
}