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<cstdlib>
#include<queue>
#include<set>
#include<vector>
#include<string>
#include<algorithm>
#include<stack>
#include<map>
#include<deque>
#include<cstdio>
using namespace std;
#define rep(i,a) for(int i=0;i<a;i++)
#define mp make_pair
#define pb push_back
#define ll __int64
//#define __int64 long long
#define P pair<int,int>
int g[111][111];
int is[111][111];
bool ok[111][111];
int n,m;
int a,b,c;
vector<P>edge;
int main(){
	cin>>n>>m;
	rep(i,m){
		cin>>a>>b>>c;
		a--;b--;
		g[a][b]=c;
		g[b][a]=c;
		is[a][b]=is[b][a]=c;
		edge.push_back(mp(a,b));
	}

	rep(k,n){
		rep(i,n){
			rep(j,n)g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
		}
	}

	rep(k,edge.size()){
		int a=edge[k].first,b=edge[k].second;
		rep(i,n){
			rep(j,n){
				if(g[i][a]+is[a][b]==g[i][b])ok[a][b]=1;
			}
		}
	}
	int ans=0;
	rep(i,n){
		for(int j=i+1;j<n;j++)if(is[i][j]&&!ok[i][j])ans++;
	}		
	cout<<ans<<endl;
	
	return 0;
}