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>
using namespace std;

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

using ll = long long;
using P = pair<ll,ll>;
using graph = vector<vector<int>>;

const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ll INF = 1LL<<60;
const ll mod = 1000000007LL;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)

ll dp[105][105];

int main(){
    int N,M;
    cin>>N>>M;
    rep(i,105){
        rep(j,105) if(i!=j) dp[i][j]=INF;
    }
    vector<vector<bool>> edge(N,vector<bool>(N));
    rep(i,M){
        ll a,b,c;
        cin>>a>>b>>c;
        dp[a-1][b-1] = c;
        dp[b-1][a-1] = c;
        edge[a-1][b-1] = true;
        edge[b-1][a-1] = true;
    }
    vector<vector<bool>> done(N,vector<bool>(N));
    rep(k,N){
        rep(i,N){
            rep(j,N) {
                if(chmin(dp[i][j],dp[i][k]+dp[k][j])) done[i][j] = true;
            }
        }
    }
    ll ans = 0;
    rep(i,N){
        rep(j,N){
            if(edge[i][j]&&done[i][j]) ans++;
        }
    }
    cout<<ans/2<<endl;
}