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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
 
#define sl(x) scanf("%ld", &x)
#define ss(x) scanf("%s", x)
#define pl(x) printf("%ld\n", x)
#define ps(x) printf("%s\n", x)
 
#define INF 100000000000
#define MOD 1000000007
 
#define MIN(a,b) (((a)<(b))?(a):(b))
#define MAX(a,b) (((a)>(b))?(a):(b))

long path[401][401];
 
void init() {
	long i, j;
	
	for(i=1;i<=400;i++) {
		for(j=1;j<=400;j++) {
			if(i == j) {
				path[i][j] = 0;
			} else {
				path[i][j] = INF;
			}
		}
	}
}

int main() {
	long N, M;
	long a[1001], b[1001], c[1001];
	long i, j, k;
	long tmp;
	long ans;
	
	init();
	
	sl(N); sl(M);
	ans = M;
	for(i=1;i<=M;i++) {
		sl(a[i]); sl(b[i]); sl(c[i]);
		path[a[i]][b[i]] = c[i];
		path[b[i]][a[i]] = c[i];
	}
	
	for(i=1;i<=N;i++)
		for(j=1;j<=N;j++)
			for(k=1;k<=N;k++)
				path[j][k] = MIN(path[j][k], path[j][i] + path[i][k]);
	
	for(k=1;k<=M;k++) {
		/*
		for(i=1;i<=N;i++) {
			if(path[i][a[k]] + c[k] == path[i][b[k]]) {
				ans--;
				break;
			}
		}
		*/
		for(i=1;i<N;i++) {
			for(j=i+1;j<=N;j++) {
				if(path[i][j] == path[i][a[k]] + path[b[k]][j] + c[k]) {
					ans--;
					goto NEXT;
				}
			}
		}
		for(i=N;i>0;i--) {
			for(j=i-1;j>=0;j--) {
				if(path[i][j] == path[i][a[k]] + path[b[k]][j] + c[k]) {
					ans--;
					goto NEXT;
				}
			}
		}
	  NEXT:;
	}
	
	pl(ans);
	
	return 0;
}