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Re:半平面交的那个直线怎么构造的?

Posted by cangratul at 2009-07-28 13:11:01 on Problem 1755
In Reply To:半平面交的那个直线怎么构造的? Posted by:fhyvenus at 2009-07-15 10:59:29
看完題的感覺就是一個3D凸包,3D的好恐怖啊,后面想了下可以轉化為2D的,方法如下:
設3斷的距離為A,B,C,S為總長,再設a=A/S,b=B/S,c=C/S,sa,sb,sc為速度:
則不等式為:
   A/sa1+B/sb1+C/sc1-(A/sa2+B/sb2+C/sc2)<0;
  由于A=a*S,B=b*S,C=c*S,a+b+c=1;帶入不等式,2邊消掉S,則變為:
  a/sa1+b/sb1+(1-a-b)/sc1-(a/sa2+b/sb2+(1-a-b)/sc2)<0;(a,b是變量)
  這樣就變成2D的了,然后就是用你的模板吧,算核的面積是否>0即可;
需要注意還要加入這3個不等式:
a>0; b>0; a+b<1;
另外注意判斷面積的時候的精度,我是if(ans<1e-16) puts("No"); else puts("Yes");
試下這組數據吧
2
1 1 10000
2 2 9999

Yes
Yes

如果還AC不了,就說明你的RP負的超越多項式的范圍了,重置吧!


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