！ HDU 4173 to the point distance of not more than 2.5, find the maximum number of points can be satisfied-simple geometry

Test instructions: There are n individuals to the party, but the condition is that the party location to his home is not more than 2.5, now you need to find the best party location for the most people to party.

Analysis:

The topic looks very difficult, how to ask scope and then contain points? In fact, a center + RADIUS does not represent a circle.

Enumerates the center of a circle with a radius of 2.5 for two points, saves it, and then uses these centers to find the distance of all points to the center of the circle, recording the number of points less than or equal to 2.5, and updating the results. 200*199/2*200 does not time out, computing is the basic math knowledge of high school. Be careful with your patience. Note that the calculated floating-point number is determined using EPS.

Code:

#include <iostream> #include <cmath> #include <algorithm> #define EPS 1e-8using namespace Std;int N,ans , cnt;double x[300],y[300];struct node{double x, y;} center[40000];d ouble dis (double a,double b,double c,double d) {return sqrt ((a-c) * (a-c) + (b-d) * (b-d));} void Getcenter (int i,int j) {Double midx= (x[i]+x[j])/2.0;double midy= (Y[i]+y[j])/2.0;double k,b;if (x[i]-x[j]==0) { Double X1=sqrt (6.25-(Midy-y[i]) * (Midy-y[i]) +x[i];d ouble x2=x[i]-sqrt (6.25-(midy-y[i)) * (Midy-y[i]); center[cnt]. X=x1,center[cnt++].y=midy;center[cnt].x=x2,center[cnt++].y=midy;} else if (y[i]-y[j]==0) {double y1=sqrt (6.25-(Midx-x[i]) * (Midx-x[i])) +y[i];d ouble y2=y[i]-sqrt (6.25-(midx-x[i)) * ( Midx-x[i]); center[cnt].x=midx,center[cnt++].y=y1;center[cnt].x=midx,center[cnt++].y=y2;}    Else{double k1= (Y[i]-y[j])/(X[i]-x[j]); k=-1.0/k1; B=midy-k*midx;double tmp1= (x[i]+k*y[i]-k*b);d ouble tmp2= (x[i]+k*y[i]-k*b) * (x[i]+k*y[i]-k*b)-(k*k+1) * (x[i]*x[i]+y [i]*y[i]+b*b-2*b*y[i]-6.25];d ouble x1= (tmp1+sqrt (TMP2))/(k*k+1);d ouble x2= (tmp1-sqRT (TMP2))/(k*k+1);d ouble y1=k*x1+b;double y2=k*x2+b;center[cnt].x=x1,center[cnt++].y=y1;center[cnt].x=x2,center[ Cnt++].y=y2;}}     int main () {while (cin>>n) {Cnt=0,ans=1;     for (int i=0;i<n;i++) cin>>x[i]>>y[i];     for (int i=0;i<n;i++) {for (int j=i+1;j<n;j++) {if (DIS (x[i],y[i],x[j],y[j]) <5.0+eps) getcenter (I,J);     }} for (int i=0;i<cnt;i++) {int tot=0;     for (int j=0;j<n;j++) {if (DIS (x[j],y[j],center[i].x,center[i].y) <2.5+eps) tot++;     } ans=max (Ans,tot);     } cout<<ans<<endl; }}

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！ HDU 4173 to the point distance of not more than 2.5, find the maximum number of points can be satisfied-simple geometry