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 <iostream>
#include <math.h>
#include <algorithm>
#include <set>
#include <map>
#include <string>
using namespace std;
typedef long long ll;
#define rep(i,s,n)for(ll i=s;i<n;i++)
#define repe(i,s,n)for(ll i=s;i<=n;i++)
#define rep_r(i,s,n)for(ll i=n-1;i>=s;i--)
#define repe_r(i,s,n)for(ll i=s;i<=n;i++)
static const ll MOD = 1e9 + 7;
static const ll MAX_L = (ll)1 << 59;

ll d[100][100] = {};
ll e[100][100] = {};
ll ans = 0;
ll n;
void wf() {
	rep(k, 0, n)rep(i, 0, n)rep(j, 0, n) {
		if (d[i][j] != MAX_L) {
			d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
		}
	}
}
int main() {
	ll m; cin >> n >> m;
	rep(i, 0, n)rep(j, 0, n) {
		if (i != j) {
			d[i][j] = MAX_L;
			e[i][j] = MAX_L;
		}
		else {
			d[i][j] = 0;
			e[i][j] = 0;
		}
	}

	rep(i, 0, m) {
		ll a, b, c;
		cin >> a >> b >> c;
		a--; b--;
		d[a][b] = c;
		d[b][a] = c;
		e[a][b] = c;
		e[b][a] = c;
	}

	wf();

	rep(i, 0, n)rep(j, 0, n) {
		if (d[i][j] != MAX_L && d[i][j] != e[i][j]) {
			ans++;
		}
	}

	cout << ans / 2 << endl;

	return 0;
}