【POJ 1981 】Circle and Points
当两个点距离小于直径时,由它们为弦确定的一个单位圆(虽然有两个圆,但是想一想知道只算一个就可以)来计算覆盖多少点。
#include <cstdio> #include <cmath> #define N 301 #define eps 1e-5 using namespace std; int n,ans,tol; double x[N],y[N],dx,dy; inline double sqr(double x) { return x*x; } inline double dis(int a,int b) { return sqrt((x[a]-x[b])*(x[a]-x[b])+(y[a]-y[b])*(y[a]-y[b])); } inline void solve() { tol=0; for(int i=1; i<=n; i++) if(dis(0,i)<=1+eps) tol++; if(tol>ans)ans=tol; } inline void get(int a,int b){ double v=atan((x[b]-x[a])/(y[a]-y[b]));//等于0时会返回inf,v=pi/2 double c=sqrt(1-sqr(dis(a,b)/2.0)); dx= c*cos(v); dy= c*sin(v); } int main() { while(scanf("%d",&n)&&n) { ans=1; for (int i=1; i<=n; i++) scanf("%lf%lf",&x[i],&y[i]); for (int i=1; i<=n; i++) for (int j=i+1; j<=n; j++) if(dis(i,j)<2){ double mx=(x[i]+x[j])/2.0,my=(y[i]+y[j])/2.0; get(i,j); x[0]=mx+dx,y[0]=my+dy; solve(); } printf("%d\n",ans); } }
┆凉┆暖┆降┆等┆幸┆我┆我┆里┆将┆ ┆可┆有┆谦┆戮┆那┆ ┆大┆始┆ ┆然┆
┆薄┆一┆临┆你┆的┆还┆没┆ ┆来┆ ┆是┆来┆逊┆没┆些┆ ┆雁┆终┆ ┆而┆
┆ ┆暖┆ ┆如┆地┆站┆有┆ ┆也┆ ┆我┆ ┆的┆有┆精┆ ┆也┆没┆ ┆你┆
┆ ┆这┆ ┆试┆方┆在┆逃┆ ┆会┆ ┆在┆ ┆清┆来┆准┆ ┆没┆有┆ ┆没┆
┆ ┆生┆ ┆探┆ ┆最┆避┆ ┆在┆ ┆这┆ ┆晨┆ ┆的┆ ┆有┆来┆ ┆有┆
┆ ┆之┆ ┆般┆ ┆不┆ ┆ ┆这┆ ┆里┆ ┆没┆ ┆杀┆ ┆来┆ ┆ ┆来┆