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>

#define INF 114514810

struct iset_t {
	int data;
	struct iset_t* next[2];
};

struct iset_t* generateSet(int data, struct iset_t* next1, struct iset_t* next2) {
	struct iset_t* generated = malloc(sizeof(struct iset_t));
	generated->data = data;
	generated->next[0] = next1;
	generated->next[1] = next2;
	return generated;
}

int N, M;
int cost[128][128];
struct iset_t* usedEdges[128][128];

char edgeUsed[1024];

void traverseSet(struct iset_t* set) {
	if (set == NULL || set->data < 0) return;
	if (set->data > 0) edgeUsed[set->data] = 1;
	traverseSet(set->next[0]);
	traverseSet(set->next[1]);
}

int main(void) {
	int i, j, k;
	int count;
	if (scanf("%d%d", &N, &M) != 2) return 1;
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) {
			cost[i][j] = INF;
			usedEdges[i][j] = NULL;
		}
		cost[i][i] = 0;
	}
	for (i = 1; i <= M; i++) {
		int a, b, c;
		if (scanf("%d%d%d", &a, &b, &c) != 3) return 1;
		cost[a][b] = cost[b][a] = c;
		usedEdges[a][b] = usedEdges[b][a] = generateSet(i, NULL, NULL);
	}
	for (k = 1; k <= N; k++) {
		for (i = 1; i <= N; i++) {
			for (j = 1; j <= N; j++) {
				int current = cost[i][j];
				int candidate = cost[i][k] + cost[k][j];
				if (candidate < current) {
					cost[i][j] = candidate;
					usedEdges[i][j] = generateSet(0, usedEdges[i][k], usedEdges[k][j]);
				} else if (candidate == current) {
					usedEdges[i][j] = generateSet(0, usedEdges[i][j],
						generateSet(0, usedEdges[i][k], usedEdges[k][j]));
				}
			}
		}
	}
	for (i = 1; i <= N; i++) {
		for (j = 1; j <= N; j++) {
			traverseSet(usedEdges[i][j]);
		}
	}
	count = 0;
	for (i = 1; i <= M; i++) {
		if (!edgeUsed[i]) count++;
	}
	printf("%d\n", count);
	return 0;
}