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 <algorithm>
using namespace std;

int N,M;
int ab[110][110], no_need[110][110];


int dik(int k){
	int now=k;
	int dis[110] = {};
	int det[110] = {};
	int prev[110][110] = {};
	for(int i=0;i<N+1;i++){
		dis[i] = 10000;
	}
	dis[k] = 0;
	det[k] = 1;

	for(int l=1;l<N;l++){
		for(int i=1;i<N+1;i++){
			if(det[i]==0 ){
				if(dis[now]+ab[now][i] < dis[i]){
					dis[i] = dis[now]+ab[now][i];
					for(int j=1;j<N+1;j++){
						prev[i][j] = 0;
					}
					prev[i][now] = 1;
				}
				if(dis[now]+ab[now][i] < dis[i]){
					prev[i][now] = 1;
				}
				
			}
		}
		now = 0;
		for(int i=1;i<N+1;i++){
			if(det[i]==0){
				if(dis[i]<dis[now]) now=i;
			}
		}
		det[now] = 1;
		for(int i=1;i<N+1;i++){
			if(prev[now][i]==1){
				no_need[now][i] = 0;
				no_need[i][now] = 0;
			}
		}
	}
	return 0;
}






int solve(){
	int ans=0;
	for(int i=1;i<N+1;i++){
		dik(i);
	}
	for(int i=1;i<N+1;i++){
		for(int j=i+1;j<N+1;j++){
			ans += no_need[i][j];
		}
	}
	cout << ans << endl;
	return 0;
}

int main(){
	int a,b,c;
	cin >> N >> M;
	for(int i=0;i<M;i++){
		cin >> a >> b >> c;
		ab[a][b] = c;
		ab[b][a] = c;
		no_need[a][b] = 1;
	}
	solve();
	return 0;
}