[线性规划 对偶 凸包 三分] Codeforces 605C #335 (Div. 1) C. Freelancer's Dreams

很显然的线性规划
对偶一下

maximum p∗x+q∗yai∗x+bi∗y<=1

这个肯定是半平面交出一个凸包然后用直线去切
直接三分就好了
有点需要精度

//ai*x+bi*y<=1 max p*x+q*y#include<cstdio>#include<cstdlib>#include<algorithm>using namespace std;typedef long double ld;const int N=100005;int n; ld p,q;ld a[N],b[N];inline ld F(ld x){  ld y=1e130;  for (int i=1;i<=n;i++)    y=min(y,(1.0-a[i]*x)/b[i]);  return p*x+q*y;}int main(){  double _p,_q,_a,_b;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  scanf("%d%lf%lf",&n,&_p,&_q),p=_p,q=_q;  for (int i=1;i<=n;i++) scanf("%lf%lf",&_a,&_b),a[i]=_a,b[i]=_b;  ld L=0,R=1e130,m1,m2;  for (int i=1;i<=n;i++) R=min(R,1.0/a[i]);  while (R-L>1e-15)    if (F(m1=L+(R-L)/3.0)<F(m2=R-(R-L)/3.0))      L=m1;    else      R=m2;  printf("%.15lf\n",(double)F((R+L)/2));  return 0;}