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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
typedef long long ll;
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#define ios ios_base::sync_with_stdio(0) ; cin.tie(0) ; cout.tie(0)
#define so sizeof
#define pb push_back
#define cl clear() ;
#define vl vector<ll>
#define sz size()
#define len length()
#define emp empty()
#define el endl;cout.flush()
#define be begin()
#define fi first
#define se second
#define br break
#define en end()
#define ro return 0
#define br break
#define eb emplace_back
#define con continue
#define ms(x) memset(x , 0, so x)
#define all(x) (x).be , (x).en
#define acc(x) accumulate((x).be , (x).en , 0ll)
#define forn(i,a,b) for(ll i=a;i<=b;++i)
#define rng_58 mt19937 rng(chrono::steady_clock::now().time_since_epoch().count())
#define vll vector<pair<ll,ll> >
#define pll pair<ll,ll>
#define plll pair<ll,pair<ll,ll>>
#define plvl pair<pair<ll,ll> ,ll>
#define mp make_pair
#define trace3(a,b,c) cerr <<"a is " << a << " b is " << b << " c is " << c << el;
#define trace4(a,b,c,d) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << el ;
#define trace5(a,b,c,d,e) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << " e is " << e << el;
#define trace6(a,b,c,d,e,f) cerr <<"a is " << a << " b is " << b << " c is " << c <<" d is " << d << " e is " << e << " f is " << f << el ;

ll dist[103][103] , ans[103][103];
ll n , m ;
const ll inf = 3e18 ;
void flloyd()
{
    for(ll k=1;k<=n;++k){
        for(ll i = 1;i<=n;++i){
            for(ll j=1;j<=n;++j){
                ans[i][j] = min(ans[i][j] , ans[i][k] + ans[k][j]);
              ans[j][i]=ans[i][j];
            }
        }
    }
}
int main()
{
   ios;
   cin >> n >> m;
   for(ll i =1 ;i<=100;++i){
    for(ll j =1;j<=100;++j){
        if(i==j) {
                dist[i][j] = 0 ;
                dist[j][i] =0 ;
                ans[i][j] = 0 ;
                ans[j][i] =  0 ;
        }
        else dist[i][j] = 3e18;
        dist[j][i] = dist[i][j];
        ans[i][j] = dist[i][j];
        ans[j][i] = ans[i][j];
    }
   }
   bool shortt[105][105] ;
   forn(i , 0 , 104 ) forn(j , 0 , 104) shortt[i][j] = false ;
   vll edges;
   forn(i , 1 , m ){
       ll x , y , d;
       cin >> x >> y >> d;
       if(x >= y) swap(x, y) ; 
       dist[x][y] = d;
       dist[y][x] = d;
       ans[x][y] = d;
       ans[y][x] = d;
       edges.pb({x , y}) ;
   }
   flloyd();
   for(auto &x : edges){

   }
   forn(i ,1 , n-1){
      forn(j , i+1 , n ){
          if(ans[i][j] >= inf) con ;
         for(auto &x : edges){
            ll s = i , ii = x.fi , jj = x.se , t = j ;
            ll d1 = ans[s][ii];
            ll d2 = dist[ii][jj] ;
            ll d3 = ans[jj][t] ;
            ll d4 = ans[s][t] ;
            if(d1 == inf or d3 == inf ) con ;
            if(d2 == inf ) con ;
            if(d4 == inf) con ;
            if((d1 + d2 + d3) == d4 and shortt[ii][jj] == false){
                shortt[ii][jj] = true;
            }
         }
      }
   }
   ll cnt = 0;
   forn(i , 1 , n ) {
      forn(j , 1 , n ) {
         if(shortt[i][j] == true) ++cnt;
      }
   }
   cout << m - cnt << el ;
   ro;
}