用 Python 写个消消乐小游戏

Posted by Python小二 on 2020-07-27

本文非首发

提到开心消消乐这款小游戏,相信大家都不陌生,其曾在 2015 年获得过玩家最喜爱的移动单机游戏奖,受欢迎程度可见一斑,本文我们使用 Python 来做个简单的消消乐小游戏。

实现

消消乐的构成主要包括三部分:游戏主体、计分器、计时器,下面来看一下具体实现。

先来看一下游戏所需 Python 库。

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import os
import sys
import time
import pygame
import random

定义一些常量,比如:窗口宽高、网格行列数等,代码如下:

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WIDTH = 400
HEIGHT = 400
NUMGRID = 8
GRIDSIZE = 36
XMARGIN = (WIDTH - GRIDSIZE * NUMGRID) // 2
YMARGIN = (HEIGHT - GRIDSIZE * NUMGRID) // 2
ROOTDIR = os.getcwd()
FPS = 30

接着创建一个主窗口,代码如下:

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pygame.init()
screen = pygame.display.set_mode((WIDTH, HEIGHT))
pygame.display.set_caption('消消乐')

看一下效果:

再接着在窗口中画一个 8 x 8 的网格,代码如下:

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screen.fill((255, 255, 220))
# 游戏界面的网格绘制
def drawGrids(self):
	for x in range(NUMGRID):
		for y in range(NUMGRID):
			rect = pygame.Rect((XMARGIN+x*GRIDSIZE, YMARGIN+y*GRIDSIZE, GRIDSIZE, GRIDSIZE))
			self.drawBlock(rect, color=(255, 165, 0), size=1
# 画矩形 block 框
def drawBlock(self, block, color=(255, 0, 0), size=2):
	pygame.draw.rect(self.screen, color, block, size)

看一下效果:

再接着在网格中随机放入各种拼图块,代码如下:

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while True:
	self.all_gems = []
	self.gems_group = pygame.sprite.Group()
	for x in range(NUMGRID):
		self.all_gems.append([])
		for y in range(NUMGRID):
			gem = Puzzle(img_path=random.choice(self.gem_imgs), size=(GRIDSIZE, GRIDSIZE), position=[XMARGIN+x*GRIDSIZE, YMARGIN+y*GRIDSIZE-NUMGRID*GRIDSIZE], downlen=NUMGRID*GRIDSIZE)
			self.all_gems[x].append(gem)
			self.gems_group.add(gem)
	if self.isMatch()[0] == 0:
		break

看一下效果:

再接着加入计分器和计时器,代码如下:

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# 显示得分
def drawScore(self):
	score_render = self.font.render('分数:'+str(self.score), 1, (85, 65, 0))
	rect = score_render.get_rect()
	rect.left, rect.top = (55, 15)
	self.screen.blit(score_render, rect)
# 显示加分
def drawAddScore(self, add_score):
	score_render = self.font.render('+'+str(add_score), 1, (255, 100, 100))
	rect = score_render.get_rect()
	rect.left, rect.top = (250, 250)
	self.screen.blit(score_render, rect)
# 显示剩余时间
def showRemainingTime(self):
	remaining_time_render = self.font.render('倒计时: %ss' % str(self.remaining_time), 1, (85, 65, 0))
	rect = remaining_time_render.get_rect()
	rect.left, rect.top = (WIDTH-190, 15)
	self.screen.blit(remaining_time_render, rect)

看一下效果:

当设置的游戏时间用尽时,我们可以生成一些提示信息,代码如下:

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while True:
	for event in pygame.event.get():
		if event.type == pygame.QUIT:
			pygame.quit()
			sys.exit()
		if event.type == pygame.KEYUP and event.key == pygame.K_r:
			flag = True
	if flag:
		break
	screen.fill((255, 255, 220))
	text0 = '最终得分: %s' % score
	text1 = '按 R 键重新开始'
	y = 140
	for idx, text in enumerate([text0, text1]):
		text_render = font.render(text, 1, (85, 65, 0))
		rect = text_render.get_rect()
		if idx == 0:
			rect.left, rect.top = (100, y)
		elif idx == 1:
			rect.left, rect.top = (100, y)
		y += 60
		screen.blit(text_render, rect)
	pygame.display.update()

看一下效果:

说完了游戏图形化界面相关的部分,我们再看一下游戏的主要处理逻辑。

我们通过鼠标来操纵拼图块,因此程序需要检查有无拼图块被选中,代码实现如下:

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def checkSelected(self, position):
	for x in range(NUMGRID):
		for y in range(NUMGRID):
			if self.getGemByPos(x, y).rect.collidepoint(*position):
				return [x, y]
	return None

我们需要将鼠标连续选择的拼图块进行位置交换,代码实现如下:

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def swapGem(self, gem1_pos, gem2_pos):
	margin = gem1_pos[0] - gem2_pos[0] + gem1_pos[1] - gem2_pos[1]
	if abs(margin) != 1:
		return False
	gem1 = self.getGemByPos(*gem1_pos)
	gem2 = self.getGemByPos(*gem2_pos)
	if gem1_pos[0] - gem2_pos[0] == 1:
		gem1.direction = 'left'
		gem2.direction = 'right'
	elif gem1_pos[0] - gem2_pos[0] == -1:
		gem2.direction = 'left'
		gem1.direction = 'right'
	elif gem1_pos[1] - gem2_pos[1] == 1:
		gem1.direction = 'up'
		gem2.direction = 'down'
	elif gem1_pos[1] - gem2_pos[1] == -1:
		gem2.direction = 'up'
		gem1.direction = 'down'
	gem1.target_x = gem2.rect.left
	gem1.target_y = gem2.rect.top
	gem1.fixed = False
	gem2.target_x = gem1.rect.left
	gem2.target_y = gem1.rect.top
	gem2.fixed = False
	self.all_gems[gem2_pos[0]][gem2_pos[1]] = gem1
	self.all_gems[gem1_pos[0]][gem1_pos[1]] = gem2
	return True

每一次交换拼图块时,我们需要判断是否有连续一样的三个及以上拼图块,代码实现如下:

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def isMatch(self):
	for x in range(NUMGRID):
		for y in range(NUMGRID):
			if x + 2 < NUMGRID:
				if self.getGemByPos(x, y).type == self.getGemByPos(x+1, y).type == self.getGemByPos(x+2, y).type:
					return [1, x, y]
			if y + 2 < NUMGRID:
				if self.getGemByPos(x, y).type == self.getGemByPos(x, y+1).type == self.getGemByPos(x, y+2).type:
					return [2, x, y]
	return [0, x, y]

当出现三个及以上拼图块时,需要将这些拼图块消除,代码实现如下:

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def removeMatched(self, res_match):
	if res_match[0] > 0:
		self.generateNewGems(res_match)
		self.score += self.reward
		return self.reward
	return 0

将匹配的拼图块消除之后,我们还需要随机生成新的拼图块,代码实现如下:

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def generateNewGems(self, res_match):
	if res_match[0] == 1:
		start = res_match[2]
		while start > -2:
			for each in [res_match[1], res_match[1]+1, res_match[1]+2]:
				gem = self.getGemByPos(*[each, start])
				if start == res_match[2]:
					self.gems_group.remove(gem)
					self.all_gems[each][start] = None
				elif start >= 0:
					gem.target_y += GRIDSIZE
					gem.fixed = False
					gem.direction = 'down'
					self.all_gems[each][start+1] = gem
				else:
					gem = Puzzle(img_path=random.choice(self.gem_imgs), size=(GRIDSIZE, GRIDSIZE), position=[XMARGIN+each*GRIDSIZE, YMARGIN-GRIDSIZE], downlen=GRIDSIZE)
					self.gems_group.add(gem)
					self.all_gems[each][start+1] = gem
			start -= 1
	elif res_match[0] == 2:
		start = res_match[2]
		while start > -4:
			if start == res_match[2]:
				for each in range(0, 3):
					gem = self.getGemByPos(*[res_match[1], start+each])
					self.gems_group.remove(gem)
					self.all_gems[res_match[1]][start+each] = None
			elif start >= 0:
				gem = self.getGemByPos(*[res_match[1], start])
				gem.target_y += GRIDSIZE * 3
				gem.fixed = False
				gem.direction = 'down'
				self.all_gems[res_match[1]][start+3] = gem
			else:
				gem = Puzzle(img_path=random.choice(self.gem_imgs), size=(GRIDSIZE, GRIDSIZE), position=[XMARGIN+res_match[1]*GRIDSIZE, YMARGIN+start*GRIDSIZE], downlen=GRIDSIZE*3)
				self.gems_group.add(gem)
				self.all_gems[res_match[1]][start+3] = gem
			start -= 1

之后反复执行这个过程,直至耗尽游戏时间,游戏结束。

最后,我们动态看一下游戏效果。

总结

本文我们使用 Python 实现了一个简单的消消乐游戏,有兴趣的可以对游戏做进一步扩展,比如增加关卡等。

源码在公号 Python小二 后台回复 200727 获取。

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