python实现俄罗斯方块

#coding:utf8#! /usr/bin/env python# 注释说明：shape表示一个俄罗斯方块形状 cell表示一个小方块import sysfrom random import choiceimport pygamefrom pygame.locals import *from block import O, I, S, Z, L, J, TCOLS = 16ROWS = 20CELLS = COLS * ROWSCELLPX = 32 # 每个cell的像素宽度POS_FIRST_APPEAR = COLS / 2SCREEN_SIZE = (COLS * CELLPX, ROWS * CELLPX)COLOR_BG = (0, 0, 0)def draw(grid, pos=None):  # grid是一个list，要么值为None，要么值为'Block'  # 非空值在eval（）的作用下，用于配置颜色  if pos: # 6x5    s = pos - 3 - 2 * COLS # upper left position    for p in range(0, COLS):      q = s + p * COLS      for i in range(q, q + 6):        if 0 <= i < CELLS:          # 0 <=i < CELLS:表示i这个cell在board内部。          c = eval(grid[i] + ".color") if grid[i] else COLOR_BG          # 执行着色。shape的cell涂对应的class设定好的颜色，否则涂黑（背景色）          a = i % COLS * CELLPX          b = i / COLS * CELLPX          screen.fill(c, (a, b, CELLPX, CELLPX))  else: # all    screen.fill(COLOR_BG)    for i, occupied in enumerate(grid):      if occupied:        c = eval(grid[i] + ".color") # 获取方块对应的颜色        a = i % COLS * CELLPX # 横向长度        b = i / COLS * CELLPX # 纵向长度        screen.fill(c, (a, b, CELLPX, CELLPX))        # fill:为cell上色, 第二个参数表示rect  pygame.display.flip()  # 刷新屏幕def phi(grid1, grid2, pos): # 4x4# 两个grid之4*4区域内是否会相撞（冲突）  s = pos - 2 - 1 * COLS # upper left position  for p in range(0, 4):    q = s + p * COLS    for i in range(q, q + 4):      try:        if grid1[i] and grid2[i]:          return False      except:        pass  return Truedef merge(grid1, grid2):  # 合并两个grid  grid = grid1[:]  for i, c in enumerate(grid2):    if c:      grid[i] = c  return griddef complete(grid):  # 减去满行  n = 0  for i in range(0, CELLS, COLS):    # 步长为一行。    if not None in grid[i:i + COLS]:    #这一句很容易理解错误。    #实际含义是：如果grid[i:i + COLS]都不是None,那么执行下面的语句      grid = [None] * COLS + grid[:i] + grid[i + COLS:]      n += 1  return grid, n#n表示减去的行数，用作统计分数pygame.init()pygame.event.set_blocked(None)pygame.event.set_allowed((KEYDOWN, QUIT))pygame.key.set_repeat(75, 0)pygame.display.set_caption('Tetris')screen = pygame.display.set_mode(SCREEN_SIZE)pygame.display.update()grid = [None] * CELLSspeed = 500screen.fill(COLOR_BG)while True: # spawn a block  block = choice([O, I, S, Z, L, J, T])()  pos = POS_FIRST_APPEAR  if not phi(grid, block.grid(pos), pos): break # you lose  pygame.time.set_timer(KEYDOWN, speed)  # repeatedly create an event on the event queue  # speed是时间间隔。。。speed越小，方块下落的速度越快。。。speed应该换为其他名字  while True: # move the block    draw(merge(grid, block.grid(pos)), pos)    event = pygame.event.wait()    if event.type == QUIT: sys.exit()    try:      aim = {        K_UNKNOWN: pos+COLS,        K_UP: pos,        K_DOWN: pos+COLS,        K_LEFT: pos-1,        K_RIGHT: pos+1,      }[event.key]    except KeyError:      continue    if event.key == K_UP:      # 变形      block.rotate()    elif event.key in (K_LEFT, K_RIGHT) and pos / COLS != aim / COLS:      # pos/COLS表示当前位置所在行      # aim/COLS表示目标位置所在行      # 此判断表示，当shape在左边界时，不允许再向左移动（越界。。）,在最右边时向右也禁止      continue    grid_aim = block.grid(aim)    if grid_aim and phi(grid, grid_aim, aim):      pos = aim    else:      if event.key == K_UP:        block.rotate(times=3)      elif not event.key in (K_LEFT, K_RIGHT):        break  grid = merge(grid, block.grid(pos))  grid, n = complete(grid)  if n:    draw(grid)    speed -= 5 * n    if speed < 75: speed = 75

#coding:utf-8#! /usr/bin/env pythonCOLS = 16ROWS = 20class Block():  color = (255,255,255)  def __init__(self):    self._state = 0  def __str__(self):    return self.__class__.__name__  def _orientations(self):    raise NotImplementedError()  def rotate(self, times=1):    for i in range(times):      if len(self._orientations())-1 == self._state:        self._state = 0        #只要_state比_orientations长度-1还要小，就让_state加1      else:        self._state += 1  def blades(self):    # 返回对应形状的一种旋转形状。(返回一个list,list中每个元素是一个(x,y))    return self._orientations()[self._state]  def grid(self, pos, cols=COLS, rows=ROWS):    # grid()函数：对于一个形状，从它的cell中的pos位置，按照orientations的位置提示，把所有cell涂色    # pos表示的是shape中的一个cell，也就是(0,0)    if cols*rows <= pos:      return None    # 这种情况应该不可能出现吧。如果出现<=的情况    # 那么，pos都跑到界外了。。    grid = [None] * cols * rows    grid[pos] = str(self)    for b in self.blades():      x, y = b      # pos/cols表示pos处于board的第几行      if pos/cols != (pos+x)/cols:        return None      i = pos + x + y * cols      if i < 0:        continue      elif cols*rows <= i:        return None      grid[i] = str(self)      # 给相应的其他位置都“涂色”,比如对于方块，是O型的，那么pos肯定是有值的，pos位于有上角。。    return grid# 以下每个形状class，_orientations（）都返回形状的列表。(0,0)一定被包含在其中，为了省略空间所以都没有写出.class O(Block):  color = (207,247,0)  def _orientations(self):    return (      [(-1,0), (-1,1), (0,1)],      )class I(Block):  color = (135,240,60)  def _orientations(self):    return (      [(-2,0), (-1,0), (1,0)],      [(0,-1), (0,1), (0,2)],      )class S(Block):  color = (171,252,113)  def _orientations(self):    return (      [(1,0), (-1,1), (0,1)],      [(0,-1), (1,0), (1,1)],      )class Z(Block):  color = (243,61,110)  def _orientations(self):    return (      [(-1,0), (0,1), (1,1)],      [(1,-1), (1,0), (0,1)],      )class L(Block):  color = (253,205,217)  def _orientations(self):    return (      [(-1,1), (-1,0), (1,0)],      [(0,-1), (0,1), (1,1)],      [(-1,0), (1,0), (1,-1)],      [(-1,-1), (0,-1), (0,1)],      )class J(Block):  color = (140,180,225)  def _orientations(self):    return (      [(-1,0), (1,0), (1,1)],      [(0,1), (0,-1), (1,-1)],      [(-1,-1), (-1,0), (1,0)],      [(-1,1), (0,1), (0,-1)],      )class T(Block):  color = (229,251,113)  def _orientations(self):    return (      [(-1,0), (0,1), (1,0)],      [(0,-1), (0,1), (1,0)],      [(-1,0), (0,-1), (1,0)],      [(-1,0), (0,-1), (0,1)],      )