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>

int a[1005];
int b[1005];
int c[1005];
int d[105][105];

int main() {

	int N, M;

	scanf("%d %d",&N,&M);

	int i,j,k;

	for (i = 0; i < M; i++) {
		scanf("%d %d %d",&a[i],&b[i],&c[i]);
		a[i]--;
		b[i]--;
	}
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			if (i == j) {
				d[i][j] = 0;
			}
			else {
				d[i][j] = 99999999;
			}
		}
	}

	for (i = 0; i < M; i++) {
		d[a[i]][b[i]] = c[i];
		d[b[i]][a[i]] = c[i];
	}

	for (k = 0; k < N; k++) {
		for (i = 0; i < N; i++) {
			for (j = 0; j < N; j++) {
				if (d[i][j] > d[i][k] + d[k][j]) {
					d[i][j] = d[i][k] + d[k][j];
				}
			}
		}
	}

	int ans = 0;

	for (i = 0; i < M; i++) {
		if (d[a[i]][b[i]] < c[i]) {
			ans++;
		}
	}
	printf("%d\n",ans);

	return 0;
}