题意:给定N个点,然后给定一个半径为R的圆,问这个圆最多覆盖多少个点。
思路:在圆弧上求扫描线。
hihocoder1508的代码。
#include<bits/stdc++.h> #define pdd pair<double,int> #define f first #define s second #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=200010; const double eps=1e-12; const double pi=acos(-1.0); struct point{ double x,y;}p[maxn]; double det(point a,point b){ return a.x*b.y-a.y*b.x;} double dot(point a,point b){ return a.x*b.x+a.y*b.y;} point operator +(point a,point b){ return point{a.x+b.x,a.y+b.y}; } point operator -(point a,point b){ return point{a.x-b.x,a.y-b.y}; } double dist(point a,point b){ return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));} int dcmp(double x){ if(fabs(x)<eps) return 0; return x>0?1:-1;} pdd a[maxn]; int ans,N; double ang1,ang2,R; void CirinterCir(point a,point b) { double L=dist(a,b)/2,t=acos(L/R); double base=atan2(b.y-a.y,b.x-a.x); ang1=base+t,ang2=base-t; } bool check(point i,point j) { double ang=(ang1+ang2)/2; point x=point{i.x+R*cos(ang),i.y+R*sin(ang)}; return dcmp(dist(x,j)-R)<=0; } void solve() { rep(i,1,N) { int cnt=1,Mx,tot=0; rep(j,1,N) { if(i==j||dist(p[i],p[j])>R+R) continue; if(p[i].x==p[j].x&&p[i].y==p[j].y) { cnt++; continue;} CirinterCir(p[i],p[j]); if(ang1>ang2) swap(ang1,ang2); if(check(p[i],p[j])){ a[++tot]=pdd(ang1-eps,-1); a[++tot]=pdd(ang2,0); } else { a[+tot]=pdd(-pi-eps,-1); a[++tot]=pdd(ang1,0); a[++tot]=pdd(ang2-eps,-1); a[++tot]=pdd(pi,0); } } sort(a+1,a+tot+1); Mx=cnt; rep(i,1,tot){ if(a[i].s==-1) cnt++; else cnt--; Mx=max(Mx,cnt); } ans=max(ans,Mx); } } int main() { scanf("%d%lf",&N,&R); rep(i,1,N) scanf("%lf%lf",&p[i].x,&p[i].y); solve(); printf("%d\n",ans); return 0; }
原文地址:https://www.cnblogs.com/hua-dong/p/11479910.html
时间: 2024-10-29 14:10:05