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<cstdio>
#include<algorithm>
using namespace std;

const int INF = 1000000000;

int main(){
	int n,m;
	int dist[100][100];
	int ans = 0;
	int a[1000],b[1000],c[1000];
	scanf("%d %d",&n,&m);
	fill(dist[0],dist[0]+10000,INF);
	for(int i=0;i<m;i++){
		scanf("%d %d %d",a+i,b+i,c+i);
		a[i]--;
		b[i]--;
		dist[a[i]][b[i]] = dist[b[i]][a[i]] = c[i];
	}
	for(int i=0;i<n;i++) dist[i][i] = 0;
	for(int k=0;k<n;k++){
		for(int i=0;i<n;i++){
			for(int j=0;j<n;j++){
				if(dist[i][k]!=INF&&dist[k][j]!=INF) dist[i][j] = min(dist[i][j],dist[i][k]+dist[k][j]);
			}
		}
	}
	for(int i=0;i<m;i++){
		bool flag = true;
		for(int s=0;s<n;s++){
			if(dist[s][a[i]]+c[i]==dist[s][b[i]]) flag = false;
		}
		if(flag) ans++;
	}
	printf("%d\n",ans);
}