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)

#define _USE_MATH_DEFINES

#include <cmath>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <limits>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <bitset>

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int dx[] = {1, 0, -1, 0, 0};
const int dy[] = {0, 1, 0, -1, 0};

const int Maxn = 110;
const int Maxm = 1010;
const int INF = (int) 1e7;

int g[Maxn][Maxn];
int a[Maxm], b[Maxm], c[Maxm];

int n, m;

int main() {
    /*
    ios::sync_with_stdio(false);
    cin.tie(0);
     */
    cin >> n >> m;
    for (int i = 0; i < n; ++i) {
        fill(g[i], g[i] + n, INF);
        g[i][i] = 0;
    }
    for (int i = 0; i < m; ++i) {
        cin >> a[i] >> b[i] >> c[i];
        a[i]--, b[i]--;
        g[a[i]][b[i]] = g[b[i]][a[i]] = c[i];
    }

    for (int k = 0; k < n; ++k) {
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (g[i][k] == INF || g[k][j] == INF) continue;
                g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
            }

        }
    }

    int cnt = 0;
    for (int i = 0; i < m; ++i) {
        bool use = false;
        for (int j = 0; j < n; ++j) {
            if (g[j][b[i]] == g[j][a[i]] + c[i]) use = true;
        }
        if (use) cnt++;
    }

    cout << m - cnt << endl;
}