"Polar sort + double pointer linear sweep" 2017 multi-school training seven HDU 6127 hard challenge

acm.hdu.edu.cn/showproblem.php?pid=6127

Test instructions

• N points in a given planar Cartesian coordinate system, where n points have a point right at each point
• This n point 22 can be connected by a line segment, defining the weights of each segment as the product of the point-to-point weights of the segments
• Now we're going to go through the origin and make a straight line, asking the line not to go through any given point.
• In the line segments of all n point 22, calculate the weights of the segments that have intersections with this line and
• Maximize this weight and output
• The topic guarantees that the given n points do not coincide and that the connection of any two points does not go through the origin.

Ideas

• A straight line through the origin divides the N points into two half planes, A, b
• It is assumed that the points in a are a1,a2....an respectively; The point weights in B are b1,b2,...... BM, respectively. The result is Suma*sumb
• Polar Sort, enumerate each starting point, linear sweep one lap, calculate the sum of the points of the half-plane, and constantly update the optimal value

"AC"

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <string>5#include <cmath>6#include <algorithm>7 8 using namespacestd;9typedefLong Longll;Ten Const intmaxn=5e4+2; One intN; A DoubleXX[MAXN]; - DoubleYY[MAXN]; - ll VAL[MAXN]; the ll ANS[MAXN]; -  - DoubleDisDoubleX1,DoubleY1,DoubleX2,Doubley2) {returnsqrt ((x1-x2) * (X1-X2) + (y1-y2) * (y1-y2));} - Const DoubleEPS = 1e-6; + struct Point - { +     Doublex; A     Doubley; at ll Val; -     Doubledis; -     DoubleAlf; - Point () {} -Point (Double_x,Double_y): X (_x), Y (_y) {} -Point (Double_x,Double_y,ll _val): X (_x), Y (_y), Val (_val) {} inPointoperator-(ConstPoint &t)Const -     { to         returnPoint (x-t.x,y-t.y); +     } -     Double operator^(ConstPoint &t)Const the     { *         return(X*T.Y)-(y*t.x); $     }Panax Notoginseng }P[MAXN]; -Point O (0.0,0.0); the BOOLcmpConstPoint &a,ConstPoint &b)//Sort by quadrant, then by polar angle, then by distance + { A     if(A.Y = =0&& B.y = =0&& a.x*b.x <=0)returnA.x>b.x; the     if(A.Y = =0&& a.x >=0&& B.Y! =0)return true; +     if(B.y = =0&& b.x >=0&& A.Y! =0)return false; -     if(B.y*a.y <=0)returnA.y>b.y; $     return((a-o) ^ (b-o)) >0.0|| (((a-o) ^ (b-o)) = =0.0&& a.x <b.x);  $ } -  - intMain () the { -     intT;Wuyiscanf"%d",&T); the      while(t--) -     { Wumemset (ans,0,sizeof(ans)); -scanf"%d",&n); Aboutll sum=0; $          for(intI=0; i<n;i++) -         { -scanf"%lf%lf%i64d",&xx[i],&yy[i],&val[i]); -p[i]=Point (Xx[i],yy[i],val[i]);  Asum+=Val[i]; +         }    theSort (p,p+n,cmp);  -         intL=1; $ ll Res; theans[0]+=p[0].val; the          for(intI=0; i<n;i++) the         {     the             if(i>0) ans[i]=ans[i-1]-p[i-1].val; -              while((p[i]-o) ^ (p[l]-o)) >0) in             { theans[i]+=P[l].val; theL= (L +1)%N;  About             }    the             if(i==0) res=ans[i]* (sum-ans[i]); the             ElseRes=max (res,ans[i]* (sum-ans[i])); the         } +printf"%i64d\n", RES);  -     } the     return 0;Bayi}

Polar sort + Linear sweep template

Template

1 BOOLcmpConstPoint &a,ConstPoint &b)//Sort by quadrant, then by polar angle, then by distance2 {3     if(A.Y = =0&& B.y = =0&& a.x*b.x <=0)returnA.x>b.x;4     if(A.Y = =0&& a.x >=0&& B.Y! =0)return true;5     if(B.y = =0&& b.x >=0&& A.Y! =0)return false;6     if(B.y*a.y <=0)returnA.y>b.y;7     return((a-o) ^ (b-o)) >0.0|| (((a-o) ^ (b-o)) = =0.0&& a.x <b.x); 8}

Polar Sort, O is the center point

1    intL=1;2 ll Res;3        //Ans[0]+=p[0].val;4          for(intI=0; i<n;i++)5         {    6             //if (i>0) ans[i]=ans[i-1]-p[i-1].val;7              while((p[i]-o) ^ (p[l]-o)) >0)8             {9                //the necessary operationsTenL= (L +1)%N;  One             }    A             //Update the answer here -}

linear sweep

"Polar sort + double pointer linear sweep" 2017 multi-school training seven HDU 6127 hard challenge