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: Time Limit Exceeded)

#float型を許すな
#numpyはpythonで
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10**7)
from collections import Counter, deque
from collections import defaultdict
from itertools import combinations, permutations, accumulate, groupby, product
from bisect import bisect_left,bisect_right
from heapq import heapify, heappop, heappush
from math import floor, ceil,pi,factorial
from operator import itemgetter
def I(): return int(input())
def MI(): return map(int, input().split())
def LI(): return list(map(int, input().split()))
def LI2(): return [int(input()) for i in range(n)]
def MXI(): return [[LI()]for i in range(n)]
def SI(): return input().rstrip()
def printns(x): print('\n'.join(x))
def printni(x): print('\n'.join(list(map(str,x))))
inf = 10**17
mod = 10**9 + 7

def warshall_floyd(d):
    #d[i][j]: iからjへの最短距離
    for k in range(n):
        for i in range(n):
            for j in range(n):
                d[i][j] = min(d[i][j],d[i][k] + d[k][j])
    return d

##############################
n,w = map(int,input().split()) #n:頂点数　w:辺の数
edge=[]
d = [[float("inf")]*n for i in range(n)]
#d[u][v] : 辺uvのコスト(存在しないときはinf)
for i in range(w):
    x,y,z = map(int,input().split())
    x-=1
    y-=1
    edge.append([x,y,z])
    d[x][y] = z
    d[y][x] = z
for i in range(n):
    d[i][i] = 0 #自身のところに行くコストは０
warshall_floyd(d)
#print(edge)
count=0
for x,y,z in edge:
    flag=True
    for i in range(n):
        for j in range(n):
            if d[i][j]==d[i][x]+z+d[y][j]:
                #print(i,x,y,j)
                flag=False
    if flag==True:
        count+=1
print(count)

#インデックスで与えられていることに注意
#ソース　https://juppy.hatenablog.com/entry/2018/11/01/蟻本_python_全点対最短経路法（ワーシャルフロイド法