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 sys
from random import choice
import pygame
from pygame.locals import *
from block import O, I, S, Z, L, J, T

COLS = 16
ROWS = 20
CELLS = COLS * ROWS
CELLPX = 32 # 每个cell的像素宽度
POS_FIRST_APPEAR = COLS / 2
SCREEN_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 True


def merge(grid1, grid2):
  # 合并两个grid
  grid = grid1[:]
  for i, c in enumerate(grid2):
    if c:
      grid[i] = c
  return grid


def 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] * CELLS
speed = 500
screen.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 python
COLS = 16
ROWS = 20

class 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)],
      )