Python OpenGL实现八号出口3D游戏复刻

以下是一个使用Python和PyOpenGL库来尝试复刻八号出口游戏的示例代码。需要注意的是，这只是一个基础的实现，可能还需要根据实际需求进行更多的优化和完善。 ```python import pygame from pygame.locals import * from OpenGL.GL import * from OpenGL.GLU import * import random # 初始化pygame pygame.init() display = (800, 600) pygame.display.set_mode(display, DOUBLEBUF | OPENGL) # 设置OpenGL视角 gluPerspective(45, (display[0] / display[1]), 0.1, 50.0) glTranslatef(0.0, 0.0, -5) # 定义过道类 class Passage: def __init__(self, exit_number): self.exit_number = exit_number self.is_abnormal = random.random() < 0.2 # 20%的概率发生异常 self.has_seen = False def draw(self): # 绘制过道 glBegin(GL_QUADS) # 地面 glColor3f(0.5, 0.5, 0.5) glVertex3f(-1, -1, -1) glVertex3f(1, -1, -1) glVertex3f(1, -1, 1) glVertex3f(-1, -1, 1) # 天花板 glColor3f(0.8, 0.8, 0.8) glVertex3f(-1, 1, -1) glVertex3f(1, 1, -1) glVertex3f(1, 1, 1) glVertex3f(-1, 1, 1) # 左墙 glColor3f(0.3, 0.3, 0.3) glVertex3f(-1, -1, -1) glVertex3f(-1, 1, -1) glVertex3f(-1, 1, 1) glVertex3f(-1, -1, 1) # 右墙 glColor3f(0.3, 0.3, 0.3) glVertex3f(1, -1, -1) glVertex3f(1, 1, -1) glVertex3f(1, 1, 1) glVertex3f(1, -1, 1) glEnd() # 绘制海报 colors = [(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 0), (1, 0, 1), (0, 1, 1), (0.5, 0.5, 0.5)] for i, color in enumerate(colors): glColor3fv(color) glBegin(GL_QUADS) glVertex3f(-0.8 + i * 0.2, 0.5, -0.9) glVertex3f(-0.6 + i * 0.2, 0.5, -0.9) glVertex3f(-0.6 + i * 0.2, 0.7, -0.9) glVertex3f(-0.8 + i * 0.2, 0.7, -0.9) glEnd() # 绘制监控摄像头 glColor3f(0.2, 0.2, 0.2) glPushMatrix() glTranslatef(0, 0.9, -0.9) glutSolidSphere(0.1, 20, 20) glPopMatrix() # 绘制穿白衬衫黑裤子的大叔 glColor3f(1, 1, 1) # 白衬衫 glBegin(GL_QUADS) glVertex3f(-0.2, -0.5, -0.9) glVertex3f(0.2, -0.5, -0.9) glVertex3f(0.2, 0.3, -0.9) glVertex3f(-0.2, 0.3, -0.9) glEnd() glColor3f(0, 0, 0) # 黑裤子 glBegin(GL_QUADS) glVertex3f(-0.2, -0.5, -0.9) glVertex3f(0.2, -0.5, -0.9) glVertex3f(0.2, -1, -0.9) glVertex3f(-0.2, -1, -0.9) glEnd() # 绘制出口指示牌 glColor3f(1, 1, 0) glBegin(GL_QUADS) glVertex3f(0.8, 0.8, -0.9) glVertex3f(0.9, 0.8, -0.9) glVertex3f(0.9, 0.9, -0.9) glVertex3f(0.8, 0.9, -0.9) glEnd() # 显示出口编号 font = pygame.font.Font(None, 36) text = font.render(str(self.exit_number), True, (255, 255, 0)) text_surface = pygame.image.tostring(text, "RGBA", True) glWindowPos2i(700, 550) glDrawPixels(text.get_width(), text.get_height(), GL_RGBA, GL_UNSIGNED_BYTE, text_surface) # 游戏主循环 current_exit = 0 passages = [Passage(i) for i in range(9)] while True: for event in pygame.event.get(): if event.type == pygame.QUIT: pygame.quit() quit() elif event.type == KEYDOWN: if event.key == K_LEFT: # 假设按左箭头回头 if passages[current_exit].is_abnormal: current_exit = 0 else: current_exit = min(current_exit + 1, 8) elif event.key == K_RIGHT: # 假设按右箭头前进 if not passages[current_exit].is_abnormal: current_exit = min(current_exit + 1, 8) else: current_exit = 0 glClear(GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT) passages[current_exit].draw() pygame.display.flip() pygame.time.wait(10) if current_exit == 8: print("成功从8号出口出去！") break ``` ### 代码说明： 1. **初始化部分**：使用`pygame`和`OpenGL`库进行初始化，设置显示窗口和OpenGL视角。 2. **过道类`Passage`**：每个过道对象包含出口编号、是否异常的标志和是否已查看的标志。`draw`方法用于绘制过道、海报、监控摄像头、大叔和出口指示牌。 3. **游戏主循环**：处理用户输入（左箭头回头，右箭头前进），根据用户操作更新当前出口编号。在每次循环中，清除屏幕并绘制当前过道，更新显示。当到达8号出口时，游戏结束。 ######[AI问答 | 1771点数解答 | 2026-02-16 08:43:46]