gpt4 book ai didi

python - 重制蛇,无法制作更多片段

转载 作者:行者123 更新时间:2023-12-01 01:10:39 24 4
gpt4 key购买 nike

我正在 pygame 中重新制作蛇,我需要让它更长。我编写了一段代码,当我吃完苹果时,它会移动蛇的第二段:

if sa == 2:
pygame.draw.rect(screen, GREEN, pygame.Rect(sx,sy,tilesize,tilesize))

在“游戏循环”的顶部我有:

sx = x
sy = y

我想添加更多段,但不知道如何做到这一点,而不必为每个可能的段创建一个函数。我想要一个代码,它可以告诉每个新的分割去哪里,或者如果有人有更好的想法(我确信他们有),请改为发布

<小时/>

这是到目前为止我的代码:

import pygame, sys, random
from pygame.locals import *
pygame.init()
movement_x = movement_y = 0
RED = (240, 0, 0)
GREEN = (0, 255, 0)
ran = [0,25,50,75,100,125,150,175,200,225,250,275,300,325,350,375,400,425,450,475,500]
ax = 0
ay = 0
x = 0
y = 0
sa = 0
sizex = 500
sizey = 500
tilesize = 25
screen = pygame.display.set_mode((sizex,sizey))
pygame.display.set_caption('Snake')
pygame.display.set_icon(pygame.image.load('images/tile.png'))
tile = pygame.image.load('images/tile.png')
tile = pygame.transform.scale(tile, (tilesize, tilesize))
x2 = 0
pag = 0
clock = pygame.time.Clock()
sx = 0
sy = 0
vel_x = 0
vel_y = 0
ap = True
while True:
sx = x
sy = y
for event in pygame.event.get():
if event.type == QUIT:
pygame.quit()
sys.exit()
for row in range(sizex):
for column in range(sizey):
screen.blit(tile,(column*tilesize, row*tilesize,tilesize,tilesize))
for event in pygame.event.get():
if event.type == KEYDOWN:
if event.key == K_UP:
vel_y = -25
vel_x = 0
elif event.key == K_DOWN:
vel_y = 25
vel_x = 0
elif event.key == K_LEFT:
vel_x = - 25
vel_y = 0
elif event.key == K_RIGHT:
vel_x= 25
vel_y = 0
elif event.key == K_y:
pag = 1
elif event.key == K_n:
pag = 2


inBounds = pygame.Rect(0, 0, sizex, sizey).collidepoint(x+vel_x, y+vel_y)
if inBounds:
y += vel_y
x += vel_x
else:
basicFont = pygame.font.SysFont(None, 48)
text = basicFont.render('Game Over! Play again? y/n', True, GREEN, RED)
textRect = text.get_rect()
textRect.centerx = screen.get_rect().centerx
textRect.centery = screen.get_rect().centery
pygame.draw.rect(screen, RED, (textRect.left - 20, textRect.top - 20, textRect.width + 40, textRect.height + 40))
screen.blit(text, textRect)
ay = -25
ax = -25
x = -25
y = -25
if pag == 1:
pag = 0
inBounds = True
x = 0
y = 0
vel_x = 0
vel_y = 0
ax = random.choice(ran)
ay = random.choice(ran)
pygame.draw.rect(screen, RED, pygame.Rect(ax,ay,tilesize,tilesize))
if pag == 2:
pygame.quit()
sys.exit()

if ap:
pygame.draw.rect(screen, RED, pygame.Rect(ax,ay,tilesize,tilesize))
if x == ax and y == ay:
pygame.draw.rect(screen, RED, pygame.Rect(ax,ay,tilesize,tilesize))
ax = random.choice(ran)
ay = random.choice(ran)
sa += 1
if sa == 2:
pygame.draw.rect(screen, GREEN, pygame.Rect(sx,sy,tilesize,tilesize))
pygame.draw.rect(screen, GREEN, pygame.Rect(x,y,tilesize,tilesize))
pygame.display.update()
clock.tick(100)

最佳答案

想象一下蠕虫是如何移动的,它抬起头,然后在整个 body 中产生运动变化。每个 body 部分都会移动到前一个部分的位置。

你的“蛇”基本上是一个矩形列表+位置。当蛇头移动时,第二个 body 矩形会移动到蛇头移动前所占据的空间中。第三部分从第二部分移入空间,第四部分移入第三部分,依此类推。

处理此问题的一种方法是将所有“蛇部分”保留在列表中,并循环遍历条目,依次更改位置。确定头部移动的方向,然后从尾部开始,将每个段移动到前一个段的位置。

这让我们得到一个简单的函数:

snake_parts = []   # /N/ Pygame rects 

def slither( new_head_coord ):
# Move each body part to the location of the previous part
# So we iterate over the tail-parts in reverse
for i in range( len( snake_parts )-1, 0, -1 ):
x, y = snake_parts[i-1].centerx, snake_parts[i-1].centery
snake_parts[i].center = ( x, y )
# Move the head
snake_parts[0].centre = new_head_coord

关于python - 重制蛇,无法制作更多片段,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54894355/

24 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com