Powered by:NEFU AB-IN

Link

1749. 阻挡广告牌 II

• 题意

  见原题

• 思路

  只能用一块矩形盖住矩形被遮住的剩余部分，那么如果x, y其中一个未被全部遮住，那么面积只能还是割草机广告牌的面积
  其他情况分类讨论即可

• 代码

  '''
  Author: NEFU AB-IN
  Date: 2022-03-30 21:45:16
  FilePath: \ACM\Acwing\1749.py
  LastEditTime: 2022-03-30 21:53:25
  '''
  
  
  def S(x1, y1, x2, y2):
      return (x2 - x1) * (y2 - y1)
  
  
  x1, y1, x2, y2 = map(int, input().split())
  X1, Y1, X2, Y2 = map(int, input().split())
  
  if X1 <= x1 and x2 <= X2 and Y1 <= y1 and y2 <= Y2:  #全覆盖
      print(0)
  elif X1 <= x1 and x2 <= X2 and Y1 <= y1 and y1 <= Y2 <= y2:  #x全覆盖，y上面未覆盖完全
      print(S(x1, y1, x2, y2) - S(x1, y1, x2, Y2))
  elif X1 <= x1 and x2 <= X2 and Y2 >= y2 and y1 <= Y1 <= y2:  #x全覆盖，y下面未覆盖完全
      print(S(x1, y1, x2, y2) - S(x1, Y1, x2, y2))
  elif Y1 <= y1 and y2 <= Y2 and X1 <= x1 and x1 <= X2 <= x2:  #y全覆盖，x右面未覆盖完全
      print(S(x1, y1, x2, y2) - S(x1, y1, X2, y2))
  elif Y1 <= y1 and y2 <= Y2 and X2 >= x2 and x1 <= X1 <= x2:  #y全覆盖，x左面未覆盖完全
      print(S(x1, y1, x2, y2) - S(X1, y1, x2, y2))
  else:
      print(S(x1, y1, x2, y2))