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>
#include <algorithm>

using namespace std;

int main(int argc, const char *argv[]) {
  int n, m;
  cin >> n >> m;

  // d[i][j]: i-jの最短距離
  // next[i][j]: i-jの最短経路中のiの次の点
  // indeces[i][j]: i-jを直接つなぐ辺のインデックス
  vector<vector<int>> d(n, vector<int>(n, -1)), next(n, vector<int>(n, -1)),
      indeces(n, vector<int>(n, -1));
  for (int i = 0; i < m; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    d[a][b] = c;
    d[b][a] = c;
    next[a][b] = b;
    next[b][a] = a;
    indeces[a][b] = i;
    indeces[b][a] = i;
  }

  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        if (d[j][i] < 0 || d[i][k] < 0) {
          continue;
        }

        int new_dist = d[j][i] + d[i][k];
        if (d[j][k] < 0 || new_dist < d[j][k]) {
          d[j][k] = new_dist;
          next[j][k] = next[j][i];
        }
      }
    }
  }

  vector<bool> used(m, false);
  for (int i = 0; i < n - 1; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (i == j) {
        continue;
      }

      // i-jの最短経路を復元
      int cur = i;
      while (cur != j) {
        if (indeces[cur][next[cur][j]] != -1) {
          used[indeces[cur][next[cur][j]]] = true;
        }
        cur = next[cur][j];
      }
    }
  }

  cout << count(used.begin(), used.end(), false) << '\n';
}