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 <bits/stdc++.h>
using namespace std;
struct G {
    #define T pair <int, int>
    int n, m;
    vector <vector <T> > g;
    vector <pair <T, int> > e;
    void read() {
        cin >> n >> m;
        g.resize(n);
        for (int i = 0; i < m; ++ i) {
            int u, v, w; cin >> u >> v >> w;
            -- u, -- v;
            e.emplace_back(T(u, v), w);
            g[u].emplace_back(v, w);
            g[v].emplace_back(u, w);
        }
    }
    vector <int> dis;
    vector <vector <int> > old;
    pair <T, int> forbid;
    void dij(int st) {
        priority_queue <T, vector <T>, greater <T> > pq;
        pq.emplace(0, st);
        dis.resize(n);
        for (int i = 0; i < n; ++ i) dis[i] = 1e9;
        dis[st] = 0;
        while (!pq.empty()) {
            T p = pq.top(); pq.pop();
            int u = p.second, d = p.first;
            for (int i = 0; i < g[u].size(); ++ i) {
                int v = g[u][i].first, w = g[u][i].second;
                if (forbid == make_pair(T(u, v), w)) continue;
                if (dis[v] > d + w) {
                    dis[v] = d + w;
                    pq.emplace(dis[v], v);
                }
            }
        }
    }
    void solve() {
        read();
        old.resize(n);
        for (int i = 0; i < n; ++ i) {
            dij(i);
            old[i] = dis;
        }
        int ans = 0;
        for (int i = 0; i < m; ++ i) {
            forbid = e[i];
            bool ok = 1;
            for (int j = 0; j < n; ++ j) {
                dij(j);
                for (int k = 0; k < n; ++ k) {
                    if (dis[k] != old[j][k]) {
                        ok = 0;
                        break;
                    }
                }
                if (!ok) break;
            }
            ans += ok;
        }
        cout << ans << endl;
    }
} g;
int main() {
    ios :: sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    g.solve();
    return 0;
}