BZOJ1946: [Ceoi2006]antenna

First, the minimum coverage circle is obtained by the stochastic increment method, which is the upper bound of the answer.

Then the two points of the answer, when the test is enumerated each point as the origin, to find the other points are included in the circle of the angle range, and then scan the line.

Time Complexity $o (tn^2\log N) $.

#include <cstdio> #include <cmath> #include <algorithm> #include <cstdlib> #define N 510using  Namespace Std;const double Pi=acos ( -1.0), Eps=1e-8;int n,k,i,j,k;double r,ansr,lim,l,r,mid;struct p{double x, y; P () {} p (double _x,double _y) {x=_x,y=_y;}}  A[n],o,anso;struct e{double X;int t; E () {} e (double _x,int _t) {x=_x,t=_t;}} e[n<<2];inline BOOL CMP (const E&a,const e&b) {return a.x<b.x;} inline double dis (const p&x,const p&y) {return sqrt ((x.x-y.x) * (x.x-y.x) + (X.Y-Y.Y) * (X.Y-Y.Y));} Inline P Center (const p&x,const P&y,const 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);}  inline bool exist (int s,double R) {int m=0;    for (int i=0;i<n;i++) if (i!=s) {if (DIS (a[s],a[i]) >r+r+eps) continue;    Double A=atan2 (a[i].y-a[s].y,a[i].x-a[s].x), B=acos (DIS (a[s],a[i])/(R+R));  E[m++]=e (a-b,1);  E[m++]=e (a+b,-1);    E[m++]=e (a-b+pi*2,1);  E[m++]=e (a+b+pi*2,-1);  } sort (e,e+m,cmp);    for (int i=0,ret=1;i<m;i++) if ((ret+=e[i].t) ==k) {ansr=r;    Anso=p (A[s].x+cos (e[i].x) *r,a[s].y+sin (e[i].x) *r);  return 1; } return 0;}  inline bool Check (double R) {for (int i=0;i<n;i++) if (exist (I,R)) return 1; return 0;}  int main () {scanf ("%d%d", &n,&k);  for (i=0;i<n;i++) scanf ("%lf%lf", &a[i].x,&a[i].y);  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) >r+eps) {O=P (      a[i].x+a[j].x)/2, (A[I].Y+A[J].Y)/2), R=dis (A[i],o);    for (k=0;k<j;k++) if (DIS (a[k],o) >r+eps) O=center (A[k],a[j],a[i]), R=dis (A[i],o);  } lim=ansr=r,anso=o;  R=ansr-eps;    while (l+eps<r) {lim/=2;    if (lim<1e-5) break;    Mid= (L+R)/2;  if (check (mid)) R=mid-eps;else L=mid+eps; } return printf ("%.6f\n%.6f%.6f", ansr,anso.x,anso.y), 0;}

　　

BZOJ1946: [Ceoi2006]antenna