hdu 3264 Open-air shopping malls 计算几何

简单二分枚举求圆的面积交

#include<cstdio>
#include<cmath>
const double eps = 1e-8;
const double pi  = acos(-1.0);
struct Point{
    double x,y;
}p[1000];
struct circle{
    Point pp;
    double r;
}cir[30];
double area[30];
int n;
double cir_area_inst(Point c1, double r1, Point c2, double r2) {            // 两圆面积交
    double a1, a2, d, ret;
    d = sqrt((c1.x-c2.x)*(c1.x-c2.x)+(c1.y-c2.y)*(c1.y-c2.y));
    if ( d > r1 + r2 - eps ) 
        return 0;
    if ( d < r2 - r1 + eps ) 
        return pi*r1*r1;
    if ( d < r1 - r2 + eps )
        return pi*r2*r2;
    a1 = acos((r1*r1+d*d-r2*r2)/2/r1/d);
    a2 = acos((r2*r2+d*d-r1*r1)/2/r2/d);
    ret = (a1-0.5*sin(2*a1))*r1*r1 + (a2-0.5*sin(2*a2))*r2*r2;
    return ret;
}
bool judge(Point p,double r){
     int i,j;
     for(i=1;i<=n;i++){
         double Are=cir_area_inst(p,r,cir[i].pp,cir[i].r);
         if(Are<area[i]) return false;
     }
     return true;
}
int main(){
    int t,i,j;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%lf%lf%lf",&cir[i].pp.x,&cir[i].pp.y,&cir[i].r);
            area[i]=pi*cir[i].r*cir[i].r/2;
        }
        Point a;
        double rans=200000;
        for(i=1;i<=n;i++){
             a=cir[i].pp;
             double l=0,r=200000,mid;
             while(fabs(l-r)>eps){
                 mid=(l+r)/2;
                 if(judge(a,mid)){
                     r=mid;
                     if(mid<rans)
                         rans=mid;
                 }
                 else l=mid;
             }
        }
        printf("%.4lf\n",rans);
    }
    return 0;
}