Hdu3264Open-air shopping mils (Binary)

Link

The circle of the enumerated umbrella has a maximum of 20 centers, because it must overlap with an existing Center.

Then we can use the second radius.

 1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set>10 using namespace std;11 #define N 2512 #define LL long long13 #define INF 0xfffffff14 const double eps = 1e-8;15 const double pi = acos(-1.0);16 const double inf = ~0u>>2;17 struct point18 {19     double x,y;20     double r,s;21     point(double x=0,double y=0):x(x),y(y) {}22 } p[N];23 int n;24 typedef point pointt;25 pointt operator -(point a,point b)26 {27     return point(a.x-b.x,a.y-b.y);28 }29 int dcmp(double x)30 {31     if(fabs(x)<eps) return 0;32     return x<0?-1:1;33 }34 double circle_area(point a,point b)35 {36     double s,d,t,t1;37     d=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));38     if(d>=a.r+b.r) s=0;39     else if(d<=fabs(a.r-b.r)) s=min(acos(-1.0)*a.r*a.r,acos(-1.0)*b.r*b.r);40     else41     {42         t=(a.r*a.r+d*d-b.r*b.r)/2.0/d;43         t1=sqrt(a.r*a.r-t*t);44         s=-d*t1+a.r*a.r*acos(t/a.r)+b.r*b.r*acos((d-t)/b.r);45     }46     return s;47 }48 int cal(double mid,point pp)49 {50     int i;51     double sum;52     pp.r = mid;53     for(i = 1 ; i <= n ; i++)54     {55         sum = circle_area(pp,p[i]);56         if(dcmp(sum-p[i].s/2)<0) return 0;57     }58     //cout<<mid<<" "<<pp.x<<" "<<pp.y<<" "<<pp.r<<endl;59     return 1;60 }61 double solve(point pp)62 {63     double rig = 20001,lef = 0,mid;64     while(rig-lef>eps)65     {66         mid = (rig+lef)/2;67         if(cal(mid,pp))68         rig = mid;69         else lef = mid;70     }71     return rig;72 }73 int main()74 {75     int t,i;76     cin>>t;77     while(t--)78     {79         scanf("%d",&n);80         for(i = 1; i <= n ; i++)81         {82             scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].r);83             p[i].s = pi*p[i].r*p[i].r;84         }85         double ans = INF;86         for(i = 1; i <= n ; i++)87         {88             ans = min(ans,solve(p[i]));89         }90         printf("%.4f\n",ans);91     }92     return 0;93 }

View code