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>
#define l_ength size
using ll = long long;
const ll infll = (1ll<<60);
const int mod = 1000000007;
using namespace std;

struct edge{
	int from;
	int to;
	ll cost;
};

int main(){
	int n, m; cin >> n >> m;
	ll dist[n][n];
	vector<edge> graph;
	for( int i = 0; i < n; ++i ){
		for( int j = 0; j < n; ++j ){
			dist[i][j] = ( i == j ) ? 0 : infll;
		}
	}
	for( int i = 0; i < m; ++i ){
		int a, b; ll c; cin >> a >> b >> c; --a; --b;
		dist[a][b] = dist[b][a] = c;
		graph.push_back( ( edge ){ a, b, c } );
	}
	for( int k = 0; k < n; ++k ){
		for( int i = 0; i < n; ++i ){
			for( int j = 0; j < n; ++j ){
				dist[i][j] = min( dist[i][j], dist[i][k]+dist[k][j] );
			}
		}
	}
	int ans = 0;
	for( auto e : graph ){
		bool flag = false;
		for( int i = 0; i < n; ++i ){
			for( int j = i+1; j < n; ++j ){
				// この条件は「i → j の最短距離に辺 e が使われる」と同値
				if( dist[i][e.from]+e.cost+dist[e.to][j] == dist[i][j] || dist[i][e.to]+e.cost+dist[e.from][j] == dist[i][j] ){
					flag = true;
				}
			}
		}
		if( !flag ) ++ans;
	}
	cout << ans << endl;
	return 0;
}