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 <iostream>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <cmath>
#include <iomanip>
#include <queue>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <cassert>
#include <cstring>
#include <climits>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll INF = 1e18;
constexpr ll MOD = 1e9 + 7;
constexpr double PI = 3.14159265358979323846;
constexpr int dx[] = {0, 0, 1, -1};
constexpr int dy[] = {1, -1, 0, 0};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m, a[1001], b[1001], c[1001];

    int dist[110][110];

    REP(i, 110)
    REP(j, 110)
    {
        dist[i][j] = 1e8;

        if (i == j)
            dist[i][j] = 0;
    }

    cin >> n >> m;
    REP(i, m)
    {
        cin >> a[i] >> b[i] >> c[i];

        dist[a[i]][b[i]] = c[i];
        dist[b[i]][a[i]] = c[i];
    }

    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
        {
            for (int j = 0; j <= n; j++)
            {
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
            }
        }
    }

    int res = m;

    for (int i = 0; i < m; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            if (dist[j][a[i]] + c[i] == dist[j][b[i]] ||
                dist[j][b[i]] + c[i] == dist[j][a[i]])
            {
                res--;
                break;
            }
        }
    }

    cout << res << endl;

    return 0;
}