poj 1981

不是啊。我就想写个傻逼题玩玩。怎么都这么自闭啊。

找个单位圆覆盖最多的点。

n^3显然。然后被卡了。

懒得优化，考虑n^2logn的做法。

地址：https://www.cnblogs.com/weeping/p/7655975.html

相当于对每个角度记录一下贡献然后排个序扫一下。

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
typedef double db;
const db eps=1e-7;
const db pi=acos(-1);
int sign(db k){
    if (k>eps) return 1; else if (k<-eps) return -1; return 0;
}
int cmp(db k1,db k2){return sign(k1-k2);}
struct point{
    db x,y;
    point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
    point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
    point operator * (db k1) const{return (point){x*k1,y*k1};}
    point operator / (db k1) const{return (point){x/k1,y/k1};}
    db abs(){return sqrt(x*x+y*y);}
    db abs2(){return x*x+y*y;}
    db dis(point k1){return ((*this)-k1).abs();}
    point unit(){db w=abs(); return (point){x/w,y/w};}
    point turn90(){return (point){-y,x};}
    bool operator < (const point k1) const{
        int a=cmp(x,k1.x);
        if (a==-1) return 1; else if (a==1) return 0; else return cmp(y,k1.y)==-1;
    }
};
int n;point p[307],l[612];
int main(){
    while (scanf("%d",&n)&&n){
        for(int i=1;i<=n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        if(n==1){
            printf("1
");
        }else{
            int ans = 1;
            for(int i=1;i<n;i++){
                int cnt = 0;
                for(int j=1;j<=n;j++){
                    if(i==j)continue;
                    db d = p[i].dis(p[j])/2;
                    if(cmp(d,1)>0)continue;
                    db alf = atan2(p[j].y-p[i].y,p[j].x-p[i].x);
                    db dlt = acos(d);
                    l[cnt++]={alf-dlt,1};
                    l[cnt++]={alf+dlt,-1};
                }
                sort(l,l+cnt);
                int res = 1;
                for(int j=0;j<cnt;j++){
                    if(cmp(l[j].y,1)==0)res++;
                    else res--;
                    ans=max(ans,res);
                }
            }
            printf("%d
",ans);
        }
    }
}