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>
using namespace std;
using ll = long long;
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define FORD(i,a,b) for(int i=a;i>=b;i--)
#define FORL(i,x) for(int i=head[x];i;i=nxt[i])
#define ALL(a)  (a).begin(),(a).end()
#define SZ(a) int((a).size())
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
#define PB push_back
#define MP make_pair
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int dist[100][100];
const int INF = pow(10,9);


int main() {
  int n,m;
  cin >> n >> m;
  int a[m],b[m],c[m];
  FOR(i,0,n-1){
    FOR(j,0,n-1){
      if(i==j) dist[i][j] = 0;
      else dist[i][j] = INF;
    }
  }
  FOR(i,0,m-1){
    cin >> a[i] >> b[i] >> c[i];
    --a[i]; --b[i];
    dist[a[i]][b[i]] = c[i];
    dist[b[i]][a[i]] = c[i];
  }
  
  FOR(k,0,n-1){
    FOR(i,0,n-1){
      FOR(j,0,n-1){
        if(i==j) continue;
        if(i==k) continue;
        if(j==k) continue;
        if(dist[i][j]>dist[i][k]+dist[k][j]){
          dist[i][j] = dist[i][k]+dist[k][j];
        }
      }
    }
  }

  int ans = 0;

  FOR(i,0,m-1){
    bool ok = false;
    FOR(j,0,n-1){
      FOR(k,0,n-1){
        if(j==k) continue;

        if(dist[j][a[i]]+c[i]+dist[b[i]][k]==dist[j][k]) {
          ok = true;
        }
      }
    }
    if(!ok) ++ans;
  }

  cout << ans << endl;

}