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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

def gen_graph(input_list):
    graph = {}

    for item in input_list:
        if not item[0] in graph:
            graph[item[0]] = {}
        if not item[1] in graph:
            graph[item[1]] = {}
        graph[item[0]][item[1]] = item[2]
        graph[item[1]][item[0]] = item[2]

    return graph


def get_lowest_cost_vertex(costs, processed):
    lowest_cost = float("inf")
    lowest_cost_node = None
    for node in costs:
        cost = costs[node]
        if cost < lowest_cost and node not in processed:
            lowest_cost = cost
            lowest_cost_node = node
    return lowest_cost_node


def get_best_routes(graph, start):
    costs = {}
    parents = {}
    processed = []

    costs[start] = 0
    parents[start] = None
    processed.append(start)

    for item in graph[start].keys():
        costs[item] = graph[start][item]
        parents[item] = start

    for item in graph.keys():
        if not (item in costs.keys() or item == start):
            costs[item] = float("inf")
            parents[item] = None

    node = get_lowest_cost_vertex(costs, processed)

    while node is not None:
        cost = costs[node]
        neighbors = graph[node]

        for n in neighbors.keys():
            new_cost = cost + neighbors[n]
            if costs[n] > new_cost:
                costs[n] = new_cost
                parents[n] = node
        processed.append(node)
        node = get_lowest_cost_vertex(costs, processed)

    routes = []
    for item in graph.keys():
        if item != start:
            route = []
            route.append(item)
            parent = parents[item]
            while 1:
                route.append(parent)
                if parent == start:
                    break
                parent = parents[parent]
            route.reverse()
            routes.append(route)
    return routes


def main(vertex_count, input_list):
    graph = gen_graph(input_list)

    routes = []
    for start in range(1, vertex_count+1):
        for route in get_best_routes(graph, start):
            routes.append(route)

    route_strs = []
    for route in routes:
        route_strs.append(''.join(list(map(str, route))))

    side_strs = []
    for item in input_list:
        side_strs.append(str(item[0]) + str(item[1]))
        side_strs.append(str(item[1]) + str(item[0]))

    count = 0
    route_strs_joined = '/'.join(route_strs)

    for side_str in side_strs:
        if route_strs_joined.find(side_str) == -1:
            count += 1

    return count / 2


if __name__ == "__main__":
    input_1 = input()
    vertex_count, sides_count = list(map(int, input_1.split()))
    input_list = []
    for _ in range(sides_count):
        vertex_1, vertex_2, distance = list(map(int, input().split()))
        input_list.append((vertex_1, vertex_2, distance))
    print(main(vertex_count, input_list))