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;

#define int long long
// #define double long double
#define FOR(i, a, b) for(ll i = (a); i < (b); ++i)
#define FORR(i, a, b) for(ll i = (a); i > (b); --i)
#define REP(i, n) for(ll i = 0; i < (n); ++i)
#define REPR(i, n) for(ll i = n; i >= 0; i--)
#define FOREACH(x, a) for(auto &(x) : (a))
#define VECCIN(x)                                                              \
    for(auto &youso_ : (x)) cin >> youso_
#define bitcnt __builtin_popcount
#define SZ(x) ((ll)(x).size())
#define fi first
#define se second
#define All(a) (a).begin(), (a).end()
#define rAll(a) (a).rbegin(), (a).rend()
template <typename T = long long> inline T IN() {
    T x;
    cin >> x;
    return (x);
}
inline void CIN() {}
template <class Head, class... Tail>
inline void CIN(Head &&head, Tail &&... tail) {
    cin >> head;
    CIN(move(tail)...);
}
#define CCIN(...)                                                              \
    char __VA_ARGS__;                                                          \
    CIN(__VA_ARGS__)
#define DCIN(...)                                                              \
    double __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define LCIN(...)                                                              \
    ll __VA_ARGS__;                                                            \
    CIN(__VA_ARGS__)
#define SCIN(...)                                                              \
    string __VA_ARGS__;                                                        \
    CIN(__VA_ARGS__)
#define Printv(v)                                                              \
    {                                                                          \
        REP(x, v.size()) { cout << v[x] << (x == v.size() - 1 ? "\n" : " "); } \
    }
template <typename T = string> inline void eputs(T s) {
    cout << s << "\n";
    exit(0);
}
template <typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val) {
    std::fill((T *)array, (T *)(array + N), val);
}
template <typename T> using PQG = priority_queue<T, vector<T>, greater<T>>;
template <typename T> using PQ = priority_queue<T>;

typedef long long ll;
typedef pair<ll, ll> PL;
typedef vector<PL> VPL;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef vector<double> VD;

const int INF = 1e9;
const int MOD = 1e9 + 7;
const ll LINF = 1e18;
const ll dx[] = {1, 0, -1, 0};
const ll dy[] = {0, 1, 0, -1};
#define PI 3.141592653589793238

ll N, M, a[1000], b[1000], c[1000];
ll d[100][100], memo[100][100];
bool need[100][100];

void warshall_floyd() {
    REP(i, N) d[i][i] = 0;
    REP(k, N) REP(i, N) REP(j, N) { d[i][j] = min(d[i][j], d[i][k] + d[k][j]); }
}

void warshall_floyd2() {
    REP(i, N) REP(j, N) {
        if(d[i][j] == memo[i][j]) {
            need[i][j] = true;
            need[j][i] = true;
        }
    }
}

signed main() {
    cin >> N >> M;
    Fill(d, LINF);
    REP(i, M) {
        cin >> a[i] >> b[i] >> c[i];
        d[--a[i]][--b[i]] = c[i];
        d[b[i]][a[i]] = c[i];
        memo[a[i]][b[i]] = c[i];
        memo[b[i]][a[i]] = c[i];
    }
    warshall_floyd();
    warshall_floyd2();
    ll ans = 0;
    REP(i, M) if(!need[a[i]][b[i]]) ans++;
    cout << ans << "\n";
}