BZOJ1338: Pku1981 Circle and Points单位圆覆盖

来源：互联网发布：matlab和python 编辑：程序博客网时间：2024/06/13 05:21

提供一个暴力的想法
极大肯定是两个点在圆周上
然后O(n3)暴力吧卡卡常数就过了

好吧。。。
其实有个O(n2lgn) 的解法主要思路就是对于每个点求包含这个点的最被包含点集
具体看爱神Blog
http://blog.csdn.net/acm_cxlove/article/details/7894310

#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>using namespace std;constdouble r=1;const double pi=acos(-1);const  double eps=1e-7;struct Circle{   double x,y;   inline void get(){scanf("%lf%lf",&x,&y);}}P[100001];#define abs(a) ((a)<0?(-(a)):(a))struct Pair{    double bg;    int ed;    inline friend bool operator  <(Pair a,Pair b){return a.bg<b.bg||(abs(a.bg-b.bg)<eps&&a.ed<b.ed);} }Cache[100001];inline double dis(Circle a,Circle b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main(){    while(true)    {       int n,tot,ans=0;       scanf("%d\n",&n);       if(n==0)return 0;       for(int i=1;i<=n;i++)P[i].get();       for(int i=1;i<=n;i++)       {           tot=0;          for(int j=1;j<=n;j++)           if(i^j)           {               double D=dis(P[i],P[j]);               if(D>2.0)continue;               double T=atan2(P[j].y-P[i].y,P[j].x-P[i].x);               T=T<0?T+2*pi:T;               double Pt=acos(D/2.0);               Cache[++tot]=(Pair){T-Pt+2*pi,false};               Cache[++tot]=(Pair){T+Pt+2*pi,true};           }          int sum=0;          sort(Cache+1,Cache+1+tot);          for(int i=1;i<=tot;i++)           if(Cache[i].ed)sum--;           else ans=max(ans,++sum);       }       printf("%d\n",ans+1);    }    return 0;}

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