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: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(v) (v).begin(), (v).end()
#define coutd(n) cout << fixed << setprecision(n)
#define ll long long int
#define vl vector<ll>
#define vi vector<int>
const int INF = 2147483647;
const ll MOD = 1000000007;

using namespace std;

struct edge {
  int to, cost;
  bool used;
};
typedef vector<vector<edge>> graph;

void warshallFloyd(int n, const graph &g, vector<vector<int>> &d) {

  REP(i, n) {
    REP(j, n) { d[i].push_back(100000000); }
  }
  REP(i, n) {
    for (auto e : g[i]) {
      d[i][e.to] = e.cost;
    }
  }
  REP(i, n) { d[i][i] = 0; }

  REP(k, n) {
    REP(i, n) {
      REP(j, n) { d[i][j] = min(d[i][j], d[i][k] + d[k][j]); }
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  graph g(n);

  REP(i, m) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    g[a].push_back({b, c, false});
    g[b].push_back({a, c, false});
  }

  vector<vector<int>> d(n);
  warshallFloyd(n, g, d);

  REP(i, n) {
    REP(j, n) {
      REP(k, g[i].size()) {
        if (d[j][i] == d[j][g[i][k].to] + g[i][k].cost) {
          g[i][k].used = true;
        }
      }
    }
  }

  int ans = 0;
  REP(i, n) {
    for (auto e : g[i]) {
      if (!e.used) {
        ans++;
      }
    }
  }
  cout << ans / 2 << endl;
}