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 Python (3.4.3) (judge verdict: Runtime Error)

def main():
	N, M = map(int, input().split())
	points = [[] for _ in range(N+1)]
	for i in range(1, M+1):
		a, b, c = map(int, input().split())
		points[a].append([b, c, i])
		points[b].append([a, c, i])
	use = set()
	def dijksta(start):
		is_visited = [False]*(N+1)
		is_visited[0] = True
		is_visited[start] = True
		stack = list(points[start])
		while stack:
			stack.sort(key=lambda x: x[1])
			use_edge = stack.pop()
			cur = use_edge[0]
			dist = use_edge[1]
			use.add(use_edge[2])
			is_visited[cur] = True
			if all(is_visited):
				break
			add_stack = list(points[cur])
			removes = []
			for i, pair in enumerate(add_stack):
				if is_visited[pair[0]]:
					removes.append(i)
					continue
				add_stack[i][1] += dist
			removes.reverse()
			for i in removes:
				add_stack.pop(i)
			stack.extend(add_stack)
	for i in range(1, M+1):
		dijksta(i)
	ans = M - len(use)
	print(ans)

if __name__ == '__main__':
    main()