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;
#define ll long long
ll n;
ll dis[1001];
bool vis[1001];
vector<pair<pair<ll, ll>, ll>>v[1001];
set<ll>st;
void dijkstra(ll k)
{
	multiset<pair<ll, pair<ll, ll> > > s;
	s.insert({0, {0, k}});
	for (ll i = 0; i <= 1000; i++)
	{
		vis[i] = 0;
		dis[i] = INT_MAX;
	}
	while (!s.empty())
	{
		pair<ll, pair<ll, ll>>p = *s.begin();
		s.erase(s.begin());
		ll wei = p.first;
		ll node = p.second.first;
		ll edge = p.second.second;

		if (vis[node])
			continue;
		vis[node] = 1;
		st.insert(edge);
		for (ll i = 0; i < v[node].size(); i++)
		{
			if (dis[node] + v[node][i].first.second < dis[v[node][i].first.first])
				{
					dis[v[node][i].first.first] = wei + v[node][i].first.second;
					s.insert({dis[v[node][i].first.first], { v[node][i].first.first, v[node][i].second}});
				}
		}
	}
}
int main()
{

	ios_base::sync_with_stdio(0);
	cout.tie(0);
	cin.tie(0);

// #ifndef ONLINE_JUDGE
// 	freopen("input.txt" , "r" , stdin);
// 	freopen("output.txt" , "w" , stdout);
// #endif

	ll  m;
	cin >> n >> m;
	for (ll i = 1; i <= m; i++)
	{
		ll a, b, c;
		cin >> a >> b >> c;
		v[a].push_back({{b, c}, i});
		v[b].push_back({{a, c}, i});
	}
	for (ll i = 1; i <= n; i++)
	{
		dijkstra(i);
	}
	cout << m + 1 - st.size() << endl;



}