Python Solution for Usaco Silver Question Wormhole Sort

I'm struggling a bit with a USACO silver question using python, .

Basically, I think my code is correct, but I can't figure out a way to make it more efficient as I simply run out of time in the 5 test cases with larger input data.

Here's my code:

import sys
sys.setrecursionlimit(200000)

#file open
fin = open("wormsort.in", "r")
fout = open("wormsort.out", "w")

#read file
temp = fin.readline().strip().split(" ")
n, m = int(temp[0]), int(temp[1])
cows = list(map(int, fin.readline().strip().split(" ")))
temp = [i for i in range(1, n+1)]

#if cows are already sorted
if cows == temp:
    fout.write(str(-1)+ "\n")
    quit()
adjacency = {i: set() for i in range(1, n + 1)}


#sorting wormhole by weight
weight = []
for i in range(m):
    temp = list(map(int, fin.readline().strip().split(" "))) #temp storage for wormhold read
    weight.append(temp[2])
    adjacency[temp[1]].add((temp[0], temp[2]))
    adjacency[temp[0]].add((temp[1], temp[2]))
weight.sort()

tvis = [0 for _ in range(n)]

def dfs(pre, bound, color): #dfs for a single component
    tvis[pre[0]-1] = color

    for i in adjacency[pre[0]]:
        if not tvis[i[0]-1] and i[1] >= bound:
            dfs(i, bound, color)


def check(k): #check if match condition given a min weight k
    counter = 0
    for i in range(1, n+1):
        counter += 1
        if tvis[i-1] == 0:
            dfs((i, 10**9), k, counter)
        else:
            continue

    for i in range(len(tvis)):
        if tvis[cows[i]-1] == tvis[i]:
            continue
        else:
            return False
    return True

high = m-1
low = 0
#binary search
while low != high:
    tvis = [0 for _ in range(n)]
    mid = (high+low)//2


    if check(weight[mid]):
        low = mid+1
    else:
        high = mid

fout.write(str(weight[low-1])+ "\n")

The idea is that since I am trying to find a maximum least weight wormhole, I could use binary search to improve efficiency compared to linear, and in the linear search use dfs for every check to see if each connected component have both the position and the cow included.