python实现俄罗斯方块

Z时代
2024-01-10
分类：综合

网上搜到一个Pygame写的俄罗斯方块（tetris），大部分看懂的前提下增加了注释，Fedora19下运行OK的

主程序：


#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

调用的模块：


#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)],
   )