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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using ll = long long;
using ull = unsigned long long;

struct S {
  int from;
  int to;
  int cost;
};

int main(int argc, char* argv[]) {
  std::cin.tie(0);
  std::ios::sync_with_stdio(false);

  /* */
  int n, m;
  cin >> n >> m;
  vector<S> adj( m );
  vector< vector<int> > dist( n, vector<int>( n, numeric_limits<int>::max() / 3 ) );

  for (int i = 0; i < m; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    --a;  --b;
    adj[i] = S{a, b, c};
    dist[a][b] = c;
    dist[b][a] = c;
  }

  for (int i = 0; i < n; ++i) dist[i][i] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        dist[i][k] = min(dist[i][k], dist[i][j] + dist[j][k]);
      }
    }
  }

  int ans = 0;
  for (auto s : adj) {
    bool found = false;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (dist[i][s.from] + s.cost + dist[s.to][j] == dist[i][j]) {
          // used
          found = true;
          break;
        }
      }

      if (found) break;
    }
    if (!found) ++ans;
  }

  cout << ans << endl;
  /* */

  return EXIT_SUCCESS;
}