POJ 2280&&hdu 1661

Test instructions: N points on a given plane, the attributes are labeled 0 and 1, then a line is found, all the points on the line are dissolved, 1 dissolves on one side, and 0 dissolves on the other. Find out the maximum number of points that can be dissolved.

Idea: Violent enumeration of each point, scan line rotation. To do the optimization first, if the side dissolves 0, then the property is 1 to do about the current enumeration line symmetry point, so that the point on the statistic side plus the point on the line is the answer. O (n2).

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <Set>6#include <stdio.h>7#include <stdlib.h>8#include <math.h>9 #defineCLC (A, B) memset (A,b,sizeof (a))Ten #defineEPS 1e-8 OnetypedefLong LongLL; A Const intMod=47000; - Const intinf=0x3f3f3f3f; - using namespacestd; the #defineN 100010 -  - structnode - { +     intX,y,r; -     Doubleang; +}v[1010],rem[1010]; A  at intcmpConst structNode A,Const structNode B) - { -     if(a.ang<B.ang) -         return 1; -     Else -         return 0; in } -  to intSIG (Doublea) + { -     if(Fabs (a) <EPS) the         return 0; *     Else if(a>0) $         return 1;Panax Notoginseng     Else -         return-1; the } +  A intCrossstructNode A,structNode B,structnode C) the { +     return(b.x-a.x) * (C.Y-A.Y)-(c.x-a.x) * (b.y-a.y); - } $  $ intMain () - { -     intN; the      while(~SCANF ("%d", &n) &&N) -     {Wuyi         intans=0; the          for(intI=0; i<n;i++) -         { Wuscanf"%d%d%d",&v[i].x,&v[i].y,&V[I].R); -rem[i]=V[i]; About         } $          for(intI=0; i<n;i++) -         { -              for(intj=0; j<n;j++) -             { Av[j]=Rem[j]; +                 if(v[j].r==1) the                 { -v[j].x=rem[i].x*2-v[j].x; $v[j].y=rem[i].y*2-v[j].y; the                 } theV[j].ang=atan2 (v[j].y-rem[i].y,v[j].x-rem[i].x); the             } theSwap (v[i],v[0]); -Sort (v+1, v+n,cmp); in              for(intj=1, t=2; J<n&&sig (V[j].ang) <=0; j + +) the             { the                 inton=2; About                  for(; T<n&&cross (v[0],v[j],v[t]) >=0; t++) the                 { the                     if(Cross (v[0],v[j],v[t]) = =0) theon++; +                 } -Ans=max (Ans,max (t-j+1, N-(t-j+1)+on )); the             }Bayi         } thecout<<ans<<Endl; the     } -     return 0; -}

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POJ 2280&&hdu 1661