微信跳一跳学习笔记

 

 

 

 

 

 

 

 

 

 最后附上大佬的代码

from __future__ import print_function, division
import os
import sys
import time
import math
import random
from PIL import Image
import subprocess


# 分数y坐标
under_game_score_y = 300
# 长按的时间系数,请自己根据实际情况调节
press_coefficient = 1.392
# 二分之一的棋子底座高度,可能要调节
piece_base_height_1_2 = 20
# 棋子的宽度,比截图中量到的稍微大一点比较安全,可能要调节
piece_body_width = 70


def set_button_position(im):
    """
    将 swipe 设置为 `再来一局` 按钮的位置
    """
    global swipe_x1, swipe_y1, swipe_x2, swipe_y2
    w, h = im.size
    left = int(w / 2)
    top = int(1584 * (h / 1920.0))
    left = int(random.uniform(left-50, left+50))
    top = int(random.uniform(top-10, top+10))    # 随机防 ban
    swipe_x1, swipe_y1, swipe_x2, swipe_y2 = left, top, left, top


def jump(distance):
    """
    跳跃一定的距离
    """
    press_time = distance * press_coefficient
    press_time = max(press_time, 200)   # 设置 200ms 是最小的按压时间
    press_time = int(press_time)
    cmd = 'adb shell input swipe {x1} {y1} {x2} {y2} {duration}'.format(
        x1=swipe_x1,
        y1=swipe_y1,
        x2=swipe_x2,
        y2=swipe_y2,
        duration=press_time
    )
    print(cmd)
    os.system(cmd)
    return press_time


def find_piece_and_board(im):
    """
    寻找关键坐标
    """
    w, h = im.size

    piece_x_sum = 0
    piece_x_c = 0
    piece_y_max = 0
    board_x = 0
    board_y = 0
    scan_x_border = int(w / 8)  # 扫描棋子时的左右边界
    scan_start_y = 0  # 扫描的起始 y 坐标
    im_pixel = im.load()
    # 以 50px 步长,尝试探测 scan_start_y
    for i in range(int(h / 3), int(h*2 / 3), 50):
        last_pixel = im_pixel[0, i]
        for j in range(1, w):
            pixel = im_pixel[j, i]
            # 不是纯色的线,则记录 scan_start_y 的值,准备跳出循环
            if pixel != last_pixel:
                scan_start_y = i - 50
                break
        if scan_start_y:
            break
    print('scan_start_y: {}'.format(scan_start_y))

    # 从 scan_start_y 开始往下扫描,棋子应位于屏幕上半部分,这里暂定不超过 2/3
    for i in range(scan_start_y, int(h * 2 / 3)):
        # 横坐标方面也减少了一部分扫描开销
        for j in range(scan_x_border, w - scan_x_border):
            pixel = im_pixel[j, i]
            # 根据棋子的最低行的颜色判断,找最后一行那些点的平均值,这个颜
            # 色这样应该 OK,暂时不提出来
            if (50 < pixel[0] < 60) \
                    and (53 < pixel[1] < 63) \
                    and (95 < pixel[2] < 110):
                piece_x_sum += j
                piece_x_c += 1
                piece_y_max = max(i, piece_y_max)

    if not all((piece_x_sum, piece_x_c)):
        return 0, 0, 0, 0
    piece_x = int(piece_x_sum / piece_x_c)
    piece_y = piece_y_max - piece_base_height_1_2  # 上移棋子底盘高度的一半

    # 限制棋盘扫描的横坐标,避免音符 bug
    if piece_x < w/2:
        board_x_start = piece_x
        board_x_end = w
    else:
        board_x_start = 0
        board_x_end = piece_x

    for i in range(int(h / 3), int(h * 2 / 3)):
        last_pixel = im_pixel[0, i]
        if board_x or board_y:
            break
        board_x_sum = 0
        board_x_c = 0

        for j in range(int(board_x_start), int(board_x_end)):
            pixel = im_pixel[j, i]
            # 修掉脑袋比下一个小格子还高的情况的 bug
            if abs(j - piece_x) < piece_body_width:
                continue

            # 修掉圆顶的时候一条线导致的小 bug,这个颜色判断应该 OK,暂时不提出来
            if abs(pixel[0] - last_pixel[0]) \
                    + abs(pixel[1] - last_pixel[1]) \
                    + abs(pixel[2] - last_pixel[2]) > 10:
                board_x_sum += j
                board_x_c += 1
        if board_x_sum:
            board_x = board_x_sum / board_x_c
    last_pixel = im_pixel[board_x, i]

    # 从上顶点往下 +274 的位置开始向上找颜色与上顶点一样的点,为下顶点
    # 该方法对所有纯色平面和部分非纯色平面有效,对高尔夫草坪面、木纹桌面、
    # 药瓶和非菱形的碟机(好像是)会判断错误
    for k in range(i+274, i, -1):  # 274 取开局时最大的方块的上下顶点距离
        pixel = im_pixel[board_x, k]
        if abs(pixel[0] - last_pixel[0]) \
                + abs(pixel[1] - last_pixel[1]) \
                + abs(pixel[2] - last_pixel[2]) < 10:
            break
    board_y = int((i+k) / 2)

    # 如果上一跳命中中间,则下个目标中心会出现 r245 g245 b245 的点,利用这个
    # 属性弥补上一段代码可能存在的判断错误
    # 若上一跳由于某种原因没有跳到正中间,而下一跳恰好有无法正确识别花纹,则有
    # 可能游戏失败,由于花纹面积通常比较大,失败概率较低
    for j in range(i, i+200):
        pixel = im_pixel[board_x, j]
        if abs(pixel[0] - 245) + abs(pixel[1] - 245) + abs(pixel[2] - 245) == 0:
            board_y = j + 10
            break

    if not all((board_x, board_y)):
        return 0, 0, 0, 0
    return piece_x, piece_y, board_x, board_y


def pull_screenshot():
    os.system('adb shell screencap -p /sdcard/1.png')
    os.system('adb pull /sdcard/1.png .')


def check_screenshot():
    # 检查获取截图的方式
    if os.path.isfile('1.png'):
        os.remove('1.png')
    pull_screenshot()
    Image.open('./1.png').load()


def main():
    check_screenshot()

    i, next_rest, next_rest_time = (0, random.randrange(3, 10),
                                    random.randrange(5, 10))
    while True:
        pull_screenshot()
        im = Image.open('./1.png')
        # 获取棋子和 board 的位置
        piece_x, piece_y, board_x, board_y = find_piece_and_board(im)
        ts = int(time.time())
        print(ts, piece_x, piece_y, board_x, board_y)
        set_button_position(im)
        jump(math.sqrt((board_x - piece_x) ** 2 + (board_y - piece_y) ** 2))
        im.close()
        i += 1
        if i == next_rest:
            print('已经连续打了 {} 下,休息 {}s'.format(i, next_rest_time))
            for j in range(next_rest_time):
                sys.stdout.write('\r程序将在 {}s 后继续'.format(next_rest_time - j))
                sys.stdout.flush()
                time.sleep(1)
            print('\n继续')
            i, next_rest, next_rest_time = (0, random.randrange(30, 100),
                                            random.randrange(10, 60))
        # 为了保证截图的时候应落稳了,多延迟一会儿,随机值防 ban
        time.sleep(random.uniform(0.9, 1.2))


if __name__ == '__main__':
    main()
大佬的代码

 

posted @ 2020-06-04 16:38  叶落知秋max  阅读(217)  评论(0编辑  收藏  举报