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 rep(i, n) for (int i = 0; i < int(n); ++i)
#define rep1(i, n) for (int i = 1; i < int(n); ++i)
#define repx(i, x, n) for (int i = int(x); i < int(n); ++i)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; --i)
#define ALL(n) begin(n), end(n)
#define IN(a, x, b) ((a) <= (x) && (x) < (b))
#define OUT(a, x, b) ((x) < (a) || (x) < (b))
typedef long long ll;
typedef long double ld;
const ll INF = 1 << 20;
const ll MOD = 1000000007;

struct Edge
{
    int to;
    int weight;
    Edge(int t, int w) : to(t), weight(w) {}
};
using Graph = vector<vector<Edge>>;

int d[110];

int main()
{
    int N, M;
    cin >> N >> M;
    Graph G(N);
    int dist[100][100], a[1010], b[1010], c[1010];
    rep(i, M)
    {
        cin >> a[i] >> b[i] >> c[i];
        a[i]--;
        b[i]--;
    }

    rep(i, N)
    {
        rep(j, N)
        {
            if (i == j)
                dist[i][j] = 0;
            else
                dist[i][j] = INF;
        }
    }
    rep(i, M)
    {
        dist[a[i]][b[i]] = min(dist[a[i]][b[i]], c[i]);
        dist[b[i]][a[i]] = min(dist[b[i]][a[i]], c[i]);
    }

    rep(k, N)
    {
        rep(i, N)
        {
            rep(j, N)
            {
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
            }
        }
    }
    int ans = M;
    rep(i, M)
    {
        int flg = false;
        rep(j, N)
        {
            if (dist[j][a[i]] + c[i] == dist[j][b[i]])
                flg = true;
        }
        if (flg)
        {
            ans--;
        }
    }
    cout << ans << endl;

    return 0;
}