最小圆覆盖

#include <cstdio>
#include <cmath>
#include <iostream>
#define eps 1e-5
#define LDB long double 
using namespace std;

  struct point{
      LDB x,y;
  }p[500001],o;

  int n;
  LDB d;

  LDB dis(point a,point b){
      return(sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)));
  }
  
  point midpoint(point a,point b){
      point ret;
      ret.x=(a.x+b.x)/2;ret.y=(a.y+b.y)/2;
    return ret;
  }
  
  point geto(point a,point b,point c){
      point mid1=midpoint(a,b),mid2=midpoint(a,c);
      point ret;
      LDB ta1=atan2(a.x-b.x,b.y-a.y),ta2=atan2(a.x-c.x,c.y-a.y),b1,b2;
      if (fabs(ta1-acos(-1)*0.5)<eps){
          b2=mid2.y-tan(ta2)*mid2.x;
          ret.x=mid1.x;ret.y=tan(ta2)*mid1.x+b2;
          return ret;
    } 
    
    if (fabs(ta2-acos(-1)*0.5)<eps){
        b1=mid1.y-tan(ta1)*mid1.x;
        ret.x=mid2.x;ret.y=tan(ta1)*mid2.x+b1;
        return ret;
    }
    
    b1=mid1.y-tan(ta1)*mid1.x;
    b2=mid2.y-tan(ta2)*mid2.x;
    ret.x=(b2-b1)/(tan(ta1)-tan(ta2)),ret.y=b1+tan(ta1)*ret.x;
    
    return ret;
  }//求三点确定的圆 
 
  int main(){
      scanf("%d",&n);
      point o;
      o=p[1];
    LDB r=0;
      
      for (register int i=2;i<=n;i++){
          if (dis(o,p[i])<r+eps) continue;
        o=midpoint(p[1],p[i]),r=dis(o,p[i]);
        
          for (register int j=2;j<i;j++){
              if (dis(o,p[j])<r+eps) continue;
              o=midpoint(p[i],p[j]),r=dis(o,p[j]);
              
                for (register int k=1;k<j;k++){
                  if (dis(o,p[k])<r+eps) continue;
                  o=geto(p[i],p[j],p[k]);
                  r=dis(p[i],o);
              }
          }
      }
    //r为圆半径，o为圆心 
  }

复杂度o(n)

转载于:https://www.cnblogs.com/zhujiangning/p/5673673.html