Main topic
Give a number of graphs, some of which are rectangles, and the rest are round. There are some points that ask how many graphs each point is in.
Ideas
The topic does not write the data range, actually is 25w. Dare O (n^2) violence? Yes, the problem is violence. You just need to deal with one-dimensional coordinates in binary and then the second dimension of violence.
CODE
#define _crt_secure_no_warnings#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 250010using namespace STD;structpoint{Doublex, y; Point (Double_,Double__): X (_), Y (__) {} point () {}BOOL operator< (ConstPoint &a)Const{returnx < a.x; }voidRead () {scanf("%LF%LF", &x,&y); }};inline DoubleCalc (ConstPoint &a,ConstPoint &b) {return sqrt((a.x-b.x) * (a.x-b.x) + (A.Y-B.Y) * (A.Y-B.Y));struct_point:point{intID;} Point[max];intAns[max];intgraphs,points;structsquare{Point P1,p2;voidRead () {DoubleX1,y1,x2,y2;scanf("%LF%LF%LF%LF", &x1,&y1,&x2,&y2); P1 = Point (min (x1,x2), Min (y1,y2)); P2 = Point (Max (X1,X2), Max (y1,y2)); }BOOLInRange (DoubleY) {returnY > P1.y && y < p2.y; }voidCount () {intStart = Upper_bound (Point +1, point + points +1, p1)-point;intEnd = Lower_bound (Point +1, point + points +1, p2)-point-1; for(inti = start; I <= end; ++i)if(InRange (POINT[I].Y)) ++ans[point[i].id]; }}square[max];structcircle{point P;DoubleRvoidRead () {p.read ();scanf("%LF", &r); }BOOLInRange (ConstPoint &a) {returnCalc (a,p) < R; }voidCount () {intStart = Upper_bound (Point +1, point + points +1, Point (P.x-r,0))-point;intEnd = Lower_bound (Point +1, point + points +1, point (p.x + R,0))-Point-1; for(inti = start; I <= end; ++i)if(InRange (Point[i])) ++ans[point[i].id]; }}circle[max];intSquares,circles;Chars[Ten];intMain () {Cin>> graphs >> points; for(inti =1; I <= graphs; ++i) {scanf('%s ', s);if(s[0] ==' R ') Square[++squares]. Read ();ElseCircle[++circles]. Read (); } for(inti =1; I <= points; ++i) {Point[i]. Read (); Point[i].id = i; } sort (Point +1, point + points +1); for(inti =1; I <= squares; ++i) Square[i]. Count (); for(inti =1; I <= circles; ++i) Circle[i]. Count (); for(inti =1; I <= points; ++i)printf("%d\n", Ans[i]);return 0;}
Bzoj 1199 Hnoi 2005 Tom's Game Computational geometry