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 (3.8.2) (judge verdict: Accepted)

import heapq


def dijkstra(s):
    dist[s] = 0
    que = []
    heapq.heappush(que, (0, s))
    while len(que) != 0:
        p = heapq.heappop(que)
        v = p[1]
        if (dist[v] < p[0]):
            continue
        for e in graph[v]:
            if dist[e[0]] > dist[v] + e[1]:
                dist[e[0]] = dist[v] + e[1]
                prev[e[0]] = v
                heapq.heappush(que, (dist[e[0]], e[0]))


def get_path(t):
    path = []
    while t != -1:
        path.append(t)
        t = prev[t]
    path = path[::-1]
    return path


N, W = map(int, input().split())
graph = [[] for i in range(N)]
edges = {}
for i in range(W):
    x, y, z = map(int, input().split())
    x -= 1
    y -= 1
    graph[x].append((y, z))
    graph[y].append((x, z))
    edges[(x, y)] = True

for st in range(N):
    dist = [float("inf")] * N
    prev = [-1] * N
    dijkstra(st)
    for i in range(N):
        gp = get_path(i)
        for i in range(1, len(gp)):
            edges[(gp[i-1], gp[i])] = False

print(sum(edges.values()))