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 (7.3.0) (judge verdict: Accepted)

def warshall_floyd(graph):
    n_nodes = len(graph)

    froms = []
    for i in range(len(graph)):
        froms.append([])
        for j in range(len(graph[i])):
            if i == j:
                froms[i].append(0)
            elif graph[i][j] < float("inf"):
                froms[i].append(i)
            else:
                froms[i].append(None)

    for k in range(n_nodes):
        for i in range(n_nodes):
            for j in range(n_nodes):
                adj_cost = graph[i][k] + graph[k][j]
                if adj_cost < graph[i][j]:
                    graph[i][j] = adj_cost
                    froms[i][j] = froms[k][j]
    
    has_negative_loop = False
    for i in range(n_nodes):
        if graph[i][i] < 0:
            has_negative_loop = True
            break

    return graph, froms, has_negative_loop

def restore_path(froms, start, goal):
    path = []
    cur = goal
    path.append(goal)
    while cur != start:
        prev = froms[start][cur]
        path.append(prev)
        cur = prev
    path.reverse()
    return path

def resolve():
    N, M = list(map(int, input().split()))
    graph = [[float("inf") for _ in range(N)] for __ in range(N)]
    import copy
    exists = [[False for _ in range(N)] for __ in range(N)]
    for i in range(M):
        a, b, c = list(map(int, input().split()))
        graph[a-1][b-1] = c
        graph[b-1][a-1] = c
        exists[a-1][b-1] = True
        exists[b-1][a-1] = True
    costs, froms, _ = warshall_floyd(graph)
    # print(froms)

    for i in range(N):
        for j in range(N):
            if i == j:
                continue
            path = restore_path(froms, i, j)
            # print("{} to {}".format(i, j))
            # print(path)
            src = 0
            for dst in range(1, len(path)):
                exists[path[src]][path[dst]] = False
                exists[path[dst]][path[src]] = False
                src = dst
    
    print(sum(sum(exists, []))//2)



if __name__ == '__main__':
    resolve()