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: Wrong Answer)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
#include <stack>
#include <functional>

#ifdef LOCAL
    #define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
    #define eprintf(...) 42
#endif

#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)
#define repp(i, m, n) for(int i = m, i##_len = (n); i < i##_len; ++i)
#define reprev(i, n) for(int i = (n-1LL); i >= 0; --i)
#define all(x) (x).begin(), (x).end()
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; }
template <class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair <int,int> P;
typedef bitset<1000> B;
typedef pair <int, bitset<1000> > D;
typedef long double ld;

const int INF = 2000000;
int main(void)
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m;
    cin >> n >> m;
    B bs;
    vector<vector<D> > mat(n, vector<D>(n, make_pair(INF, bs)));
    rep (i, n) {
        mat[i][i] = make_pair(0, bs);
    }
    rep (i, m) {
        int a, b, c;
        cin >> a >> b >> c;
        a--; b--;
        B tmp;
        tmp.set(i);
        mat[a][b] = mat[b][a] = make_pair(c, tmp);
    }
    rep (k, n) {
        rep (i, n) {
            rep (j, n) {
                if (mat[i][k].first == INF || mat[k][j].first == INF) continue;
                if (chmax(mat[i][j].first, mat[i][k].first + mat[k][j].first)) {
                    mat[i][j].second = mat[i][k].second | mat[k][j].second;
                }
            }
        }
    }
    rep (i, n) {
        rep (j, n) {
            bs |= mat[i][j].second;
        }
    }
    cout << m - bs.count() << "\n";
    return 0;
}