POJ - 1981 ：Circle and Points （圆的扫描线） hihocoder1508

Posted 2020-11-12 hua-dong

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
",ans);
    return 0;
}