Codeforces Round #514 (Div. 2) D. Nature Reserve

Http://codeforces.com/contest/1059/problem/D

Maximum value:

The bottom left and right bottom have one point respectively

R^2-(r-1) ^2 = (10^7) ^2

Maxr<0.5*10^14

Way1:

Two points.

Difference:

If you use 5*10^13-10^-6,2^ 60~70 interval, the pow,sqrt operation, the measured timeout.

is actuallyUse.

Time of a case:->

Code:

1#include <bits/stdc++.h>2 using namespacestd;3 Const intmaxn=1e5+Ten;4 5 DoubleX[MAXN],Y[MAXN];6 Doublez=5.0*1e13;7 intN;8 9 BOOLWorkDoublem)Ten { One     Doublep,q,d; A     inti; -P=-z; q=Z; -      for(i=1; i<=n;i++) the     { -D=sqrt (Pow (M,2)-pow (M-y[i],2)); -P=max (p,x[i]-d); -Q=min (q,x[i]+d); + //if (p>q) - //return 0; +     } A //return 1; at     if(p<=q) -         return 1; -     Else -         return 0; - } -  in intMain () - { to     DoubleL=0, r,m; +     intv,i; -scanf"%d",&n); thev=0; *      for(i=1; i<=n;i++) $     {Panax Notoginsengscanf"%LF%LF",&x[i],&y[i]); -         if(y[i]!=0) the         { +             if(v==0) Av= (y[i]>0); the             Else if(V!= (y[i]>0)) +             { -printf"-1"); $                 return 0; $             } -         } -y[i]=fabs (Y[i]); theL=max (l,y[i]/2); -     }WuyiR=Z; the      while((r-l)/max (1.0, l) >1e-6) -     { WuM= (L+R)/2; -         if(Work (m)) AboutR=m; $         Else -L=m; -     } -printf"%.10f", R); A     return 0; +}

Way2:

Multiple two functions F1,f2,..., FN,

F (x) =max (F1 (x), F2 (x),..., fn (x)),

And F is first reduced, then enlarged, using a three-point method

Similar:

hdu4717 The Moving Points

Codeforces Round #514 (Div. 2) D. Nature Reserve