hdu3264Open-air shopping malls(二分）

链接

枚举伞的圆心，最多只有20个，因为必须与某个现有的圆心重合。

然后再二分半径就可以了。

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 25
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 struct point
 {
     double x,y;
     double r,s;
     point(double x=,double y=):x(x),y(y) {}
 } p[N];
 int n;
 typedef point pointt;
 pointt operator -(point a,point b)
 {
     return point(a.x-b.x,a.y-b.y);
 }
 int dcmp(double x)
 {
     if(fabs(x)<eps) return ;
     return x<?-:;
 }
 double circle_area(point a,point b)
 {
     double s,d,t,t1;
     d=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
     if(d>=a.r+b.r) s=;
     else if(d<=fabs(a.r-b.r)) s=min(acos(-1.0)*a.r*a.r,acos(-1.0)*b.r*b.r);
     else
     {
         t=(a.r*a.r+d*d-b.r*b.r)/2.0/d;
         t1=sqrt(a.r*a.r-t*t);
         s=-d*t1+a.r*a.r*acos(t/a.r)+b.r*b.r*acos((d-t)/b.r);
     }
     return s;
 }
 int cal(double mid,point pp)
 {
     int i;
     double sum;
     pp.r = mid;
     for(i =  ; i <= n ; i++)
     {
         sum = circle_area(pp,p[i]);
         if(dcmp(sum-p[i].s/)<) return ;
     }
     //cout<<mid<<" "<<pp.x<<" "<<pp.y<<" "<<pp.r<<endl;
     return ;
 }
 double solve(point pp)
 {
     double rig = ,lef = ,mid;
     while(rig-lef>eps)
     {
         mid = (rig+lef)/;
         if(cal(mid,pp))
         rig = mid;
         else lef = mid;
     }
     return rig;
 }
 int main()
 {
     int t,i;
     cin>>t;
     while(t--)
     {
         scanf("%d",&n);
         for(i = ; i <= n ; i++)
         {
             scanf("%lf%lf%lf",&p[i].x,&p[i].y,&p[i].r);
             p[i].s = pi*p[i].r*p[i].r;
         }
         double ans = INF;
         for(i = ; i <= n ; i++)
         {
             ans = min(ans,solve(p[i]));
         }
         printf("%.4f\n",ans);
     }
     return ;
 }