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提到开心消消乐这款小游戏，置信大家都不生疏，其曾在 2015 年取得过玩家最青睐的挪动单机游戏奖，受欢迎水平可见一斑，本文咱们应用 Python 来做个简略的消消乐小游戏。
实现
消消乐的形成次要包含三局部：游戏主体、计分器、计时器，上面来看一下具体实现。
先来看一下游戏所需 Python 库。
import os
import sys
import time
import pygame
import random定义一些常量，比方：窗口宽高、网格行列数等，代码如下：
WIDTH = 400
HEIGHT = 400
NUMGRID = 8
GRIDSIZE = 36
XMARGIN = (WIDTH - GRIDSIZE * NUMGRID) // 2
YMARGIN = (HEIGHT - GRIDSIZE * NUMGRID) // 2
ROOTDIR = os.getcwd()
FPS = 30接着创立一个主窗口，代码如下：
pygame.init()
screen = pygame.display.set_mode((WIDTH, HEIGHT))
pygame.display.set_caption('消消乐')看一下成果：
再接着在窗口中画一个 8 x 8 的网格，代码如下：
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)看一下成果：
再接着在网格中随机放入各种拼图块，代码如下：
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看一下成果：
再接着退出计分器和计时器，代码如下：
# 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)看一下成果：
当设置的游戏工夫用尽时，咱们能够生成一些提示信息，代码如下：
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 键从新开始'
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()看一下成果：
说完了游戏图形化界面相干的局部，咱们再看一下游戏的次要解决逻辑。
咱们通过鼠标来操纵拼图块，因而程序须要查看有无拼图块被选中，代码实现如下：
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咱们须要将鼠标间断抉择的拼图块进行地位替换，代码实现如下：
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每一次替换拼图块时，咱们须要判断是否有间断一样的三个及以上拼图块，代码实现如下：
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]当呈现三个及以上拼图块时，须要将这些拼图块打消，代码实现如下：
def removeMatched(self, res_match):
if res_match[0] > 0:
self.generateNewGems(res_match)
self.score += self.reward
return self.reward
return 0将匹配的拼图块打消之后，咱们还须要随机生成新的拼图块，代码实现如下：
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之后重复执行这个过程，直至耗尽游戏工夫，游戏完结。
最初，咱们动静看一下游戏成果。
总结
本文咱们应用 Python 实现了一个简略的消消乐游戏，有趣味的能够对游戏做进一步扩大，比方减少关卡等。
源码在公众号 Python 小二 后盾回复 200727 获取。
申明：本文作者为自己，但非首发于集体号
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