bzoj1946: [Ceoi2006]ANTENNA

Posted 2020-10-01 ccz181078

给出N个点的坐标，要求能够覆盖其中至少K个点的圆的最小半径及圆心位置。

最优答案圆周上一定有点，枚举这个点，二分半径扫描线判定，随机打乱输入顺序并加上最优性剪枝，由于答案期望变优$O(logn)$次，期望时间复杂度为$O(n^2logn+nlognlog(\frac{MaxAns}{eps}))$

#include<bits/stdc++.h>
const double pi=acos(-1);
int n,k;
struct pos{
    int x,y;
    double a,d;
    void cal(){
        a=atan2(y,x);
        d=sqrt(x*x+y*y);
    }
    pos operator-(pos w)const{return (pos){x-w.x,y-w.y};}
    bool operator<(pos w)const{return a<w.a;}
}ps[555],as[555],p0;
int ap;
double ans=20000,ax=5000,ay=5000;
struct ev{
    double a;
    int t;
    bool operator<(ev w)const{return a<w.a;}
}e[1111];
void upd(double r,double a){
    if(r<ans){
        ans=r;
        ax=p0.x+cos(a)*r;
        ay=p0.y+sin(a)*r;
    }
}
bool chk(double R){
    double R2=R*2;
    int ep=0,s=0;
    for(int i=0;i<ap;++i)if(as[i].d<=R2){
        double b=acos(as[i].d/R2),l=as[i].a-b,r=as[i].a+b;
        if(l<-pi)l+=pi*2;
        if(r>pi)r-=pi*2;
        s+=l>r;
        e[ep++]=(ev){l,1};
        e[ep++]=(ev){r,-1};
    }
    if(s>=k){
        upd(R,-pi);
        return 1;
    }
    std::sort(e,e+ep);
    for(int i=0;i<ep;++i)if((s+=e[i].t)>=k){
        upd(R,e[i].a);
        return 1;
    }
    return 0;
}
int main(){
    scanf("%d%d",&n,&k);
    --k;
    for(int i=0,x,y;i<n;++i){
        scanf("%d%d",&x,&y);
        ps[i]=(pos){x,y};
    }
    std::random_shuffle(ps,ps+n);
    for(int i=0;i<n;++i){
        ap=0;
        p0=ps[i];
        for(int j=0;j<n;++j){
            as[j]=ps[j]-p0;
            as[j].cal();
            if(as[j].x|as[j].y)as[ap++]=as[j];
        }
        std::sort(as,as+ap);
        double L=0,R=ans,M=ans-5e-9;
        while(R-L>1e-8){
            if(chk(M))R=M;
            else L=M;
            M=L+(R-L)/2;
        }
    }
    printf("%.6f\n%.6f %.6f",ans,ax,ay);
    return 0;
}