不嘻嘻，没见过这种题，需要把这个红线还原重组成二维码，搜索一个是这个Peano曲线

from PIL import Image
from tqdm import tqdm


def peano(n):
    if n == 0:
        return [[0, 0]]
    else:
        in_lst = peano(n - 1)
        lst = in_lst.copy()
        px, py = lst[-1]
        lst.extend([px - i[0], py + 1 + i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px + i[0], py + 1 + i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px + 1 + i[0], py - i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px - i[0], py - 1 - i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px + i[0], py - 1 - i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px + 1 + i[0], py + i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px - i[0], py + 1 + i[1]] for i in in_lst)
        px, py = lst[-1]
        lst.extend([px + i[0], py + 1 + i[1]] for i in in_lst)
        return lst


order = peano(6)

img = Image.open(r"F:\WeChat Files\wxid_zzy3l6ymu9oh22\FileStorage\File\2024-10\网鼎杯\misc4\1.png")

width, height = img.size

block_width = width  # // 3
block_height = height  # // 3

new_image = Image.new("RGB", (width, height))

for i, (x, y) in tqdm(enumerate(order)):
    # 根据列表顺序获取新的坐标
    new_x, new_y = i % width, i // width
    # 获取原图像素
    pixel = img.getpixel((x, height - 1 - y))
    # 在新图像中放置像素
    new_image.putpixel((new_x, new_y), pixel)

new_image.save("rearranged_image.jpg")

以后多积累点图片的脚本吧，不嘻嘻