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: Accepted)

import sys
MAX_INT = int(10e10)
MIN_INT = -MAX_INT
mod = 1000000007
sys.setrecursionlimit(1000000)
def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(sys.stdin.readline())
def S(): return input()

N, M = IL()
abc = [IL() for _ in range(M)]

dp = [[MAX_INT]*N for _ in range(N)]
for a, b, c in abc:
  dp[a - 1][b - 1] = c
  dp[b - 1][a - 1] = c
for i in range(N):
  dp[i][i] = 0

for k in range(N):
  for i in range(N):
    for j in range(N):
      dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j])
#print(dp)

used = [False]*M
cnt = 0
for edge in range(M):
  a, b, c = abc[edge]
  for i in range(N):
    for j in range(N):
      if dp[i][a - 1] + c + dp[b - 1][j] == dp[i][j]:
        used[edge] = True
        cnt += 1
        break
    else:
      continue
    break

#print(used)
print(M - cnt)