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 osimport sysimport timeimport pygameimport random定义一些常量，比方：窗口宽高、网格行列数等，代码如下：
WIDTH = 400HEIGHT = 400NUMGRID = 8GRIDSIZE = 36XMARGIN = (WIDTH - GRIDSIZE * NUMGRID) // 2YMARGIN = (HEIGHT - GRIDSIZE * NUMGRID) // 2ROOTDIR = 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