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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <complex>
using namespace std;

using ll = long long;
using ull = uint64_t;
using P = pair<int, int>;
constexpr double EPS = 1e-9;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;

template<class T>
inline bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;
    vector<tuple<int, int, int>> es;
    vector<vector<int>> dist(n, vector<int>(n, INF));
    for(int i = 0; i < m; ++i){
        int a, b, c;
        cin >> a >> b >> c;
        --a, --b;
        es.emplace_back(a, b, c);
        dist[a][b] = dist[b][a] = c;
    }
    for(int i = 0; i < n; ++i)  dist[i][i] = 0;
    
    for(int k = 0; k < n; ++k){
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                chmin(dist[i][j], dist[i][k] + dist[k][j]);
            }
        }
    }

    int ans = 0;
    for(auto e : es){
        int a, b, c;
        tie(a, b, c) = e;
        bool ok = false;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                if(dist[i][j] == dist[i][a] + dist[b][j] + c){
                    ok = true;
                    break;
                }
                if(dist[i][j] == dist[i][b] + dist[a][j] + c){
                    ok = true;
                    break;
                }
            }
            if(ok)  break;
        }
        if(ok)  ++ans;
    }
    ans = m - ans;
    cout << ans << endl;
}