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: Wrong Answer)

import heapq
n, m = map(int, input().split())
g = [[] for _ in range(n)]
for i in range(m):
    a, b, c = map(int, input().split())
    g[a-1].append((b-1,c))
    g[b-1].append((a-1,c))
vs = [[False]*n for _ in range(n)]
for i in range(n):
    ds = [-1]*n
    hq = [(0,i,None)]
    while hq:
        c,idx,v = heapq.heappop(hq)
        if ds[idx] != -1:
            continue
        ds[idx] = c
        if v != None:
            vs[v[0]][v[1]] = True
        for j in g[idx]:
            if ds[j[0]] == -1:
                heapq.heappush(hq,(c+j[1],j[0],(idx,j[0])))
            elif ds[j[0]] == c+j[1]:
                vs[idx][j[0]] = True
res = 0
for i in range(n):
    for j in g[i]:
        if vs[i][j[0]] == False:
            res += 1
print(res//2)