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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

typedef enum {
  false,
  true
}boolean;

typedef struct {
	int a;
	int b;
	int c;
}edge;

typedef struct vsub{
	int v;
	int num;
	struct vsub **nexts;
	int *c;
}vertex;

boolean search(int n, vertex *ver, int sum, int dest, int border){
	if(sum >= border || n == 0){
		return false;
	}
	if(sum < border && ver->v == dest){
		return true;
	}
	int i;
	for(i = 0; i < ver->num ;i++){
		if(search(n - 1 , (ver->nexts)[i], sum + (ver->c)[i], dest, border) == true){
			return true;
		}
	}
	return false;
}

int main(){
	int N, M, i, a, b, c, ans = 0;
	scanf("%d%d", &N, &M);
	edge E[M];
	vertex *V[N];
	for(i = 0; i < N; i++){
		V[i] = (vertex *)malloc(sizeof(vertex));
		V[i]->v = i;
		V[i]->num = 0;
		V[i]->nexts = (vertex **)malloc(sizeof(vertex *) * N);
		V[i]->c = (int *)malloc(sizeof(int) * N);
	}
	for(i = 0; i < M; i++){
		scanf("%d%d%d", &a, &b, &c);
		a--;
		b--;
		E[i].a = a;
		E[i].b = b;
		E[i].c = c;
		(V[a]->nexts)[V[a]->num] = V[b];
		(V[a]->c)[V[a]->num] = c;
		(V[a]->num)++;
		(V[b]->nexts)[V[b]->num] = V[a];
		(V[b]->c)[V[b]->num] = c;
		(V[b]->num)++;
	}
	for(i = 0; i < M; i++){
		if(search(N, V[E[i].a], 0, E[i].b, E[i].c) == true){
			ans++;
		}
	}
	printf("%d\n", ans);
	return 0;
}