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 <vector>
#include <iostream>
#include <string>
#include <map>
#include <algorithm>
#include <set>
#include <queue>
#include <cstdio>
#include <utility>
#include <bitset>
#include <complex>
#include <stack>
#include <tuple>

using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef tuple<ll,ll,ll> T;

const int INF = 1e7;
int main(){
	int N,M;
	cin >> N >> M;
	vector<vector<int>> dist(N,vector<int>(N,INF));
	vector<T> in(M);
	for(int i=0;i<M;i++){
		int a,b,c;
		cin >> a >> b >> c;
		a--; b--;
		dist[a][b]=c;
		dist[b][a]=c;
		in[i]=T(a,b,c);
	}
	for(int k=0;k<N;k++){
		for(int i=0;i<N;i++)
			for(int j=0;j<N;j++)
				if(i==j)
					dist[i][j]=0;
				else
					dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
	}
	int ans=0;
	for(int i=0;i<M;i++){
		int a,b,c;
		tie(a,b,c)=in[i];
		if(dist[a][b]!=c)
			ans++;
	}
	cout << ans << endl;
	return 0;
}