BZOJ3564 : [SHOI2014]信号增幅仪

先把所有点绕原点逆时针旋转(360-a)度，再把所有点横坐标除以放大倍数p,最后用随机增量法求最小圆覆盖即可。

时间复杂度期望$O(n)$

 

#include<cstdio>

#include<algorithm>

#include<cmath>

using namespace std;

struct P{double x,y;}a[50005],o;

inline double dis(P x,P y){return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));}

inline P center(P x,P y,P z){

  double a1=y.x-x.x,b1=y.y-x.y,c1=(a1*a1+b1*b1)/2,a2=z.x-x.x,b2=z.y-x.y,c2=(a2*a2+b2*b2)/2,d=a1*b2-a2*b1;

  return (P){x.x+(c1*b2-c2*b1)/d,x.y+(a1*c2-a2*c1)/d};

}

double r,angle,e,x,y,eps=1e-8;

int n,i,j,k;

int main(){

  srand(199745);

  for(scanf("%d",&n);i<n;i++)scanf("%lf%lf",&a[i].x,&a[i].y);

  for(scanf("%lf%lf",&angle,&e),angle=(360-angle)*acos(-1.0)/180,i=0;i<n;i++)x=a[i].x,y=a[i].y,a[i].x=(cos(angle)*x-sin(angle)*y)/e,a[i].y=cos(angle)*y+sin(angle)*x;

  for(i=0;i<n;i++)swap(a[rand()%n],a[i]);

  for(o=a[0],i=1;i<n;i++)if(dis(a[i],o)>=r+eps)for(o=a[i],r=0,j=0;j<i;j++)if(dis(a[j],o)+eps>=r)for(o=(P){(a[i].x+a[j].x)/2,(a[i].y+a[j].y)/2},r=dis(o,a[i]),k=0;k<j;k++)if(dis(a[k],o)>=r-eps)o=center(a[k],a[j],a[i]),r=dis(o,a[i]);

  printf("%.3lf",r);

  return 0;

}