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>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define eps LDBL_EPSILON
#define moder 1000000007
#define int long long
#define ll long long
#define INF LLONG_MAX/3
#define P std::pair<int,int>
#define prique priority_queue<int,vector<int>,greater<int>>
using namespace std;
int n,m,a,b,c;
int d[110][110];
set<P> st,h;
signed main(){
	cin>>n>>m;
	rep(i,110){
		rep(j,110)d[i][j]=INF;
	}
	rep(i,m){
		cin>>a>>b>>c;
		d[a][b]=c;
		d[b][a]=c;
		h.insert({a,b});
		h.insert({b,a});
	}
	for(int k=1;k<=n;k++){
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				if(d[i][j]>=d[i][k]+d[k][j]){
					if(h.find({i,k})!=h.end()){
						st.insert({i,k});
						st.insert({k,i});
					}
					if(h.find({j,k})!=h.end()){
						st.insert({k,j});
						st.insert({j,k});
					}
					d[i][j]=d[i][k]+d[k][j];
				}
			}
		}
	}
	cout<<m-st.size()/2<<endl;
    return 0;
}