Open-air shopping malls（hdu 3264 二分+求两圆相交面积

 题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3264

 

 题解：每个圆作为大圆圆心遍历过去

求每个圆心 满足条件的最小半径  在所有最小半径里找min

#include<bits/stdc++.h>
using namespace std;
#define pi acos(-1.0)
double x[22],y[22],r[22];
/*1
2
0 0 1
2 0 1
*/
int n;
double mj(double x1,double y1,double r1,double x2,double y2,double r2)//相交面积 ，有模板的，刚开始我还自己写 错的
{//一开始返回成bool了
    double d = sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));
    if (d >= r1+r2)
        return 0;
    if (r1>r2)
    {
        double tmp = r1;
        r1 = r2;
        r2 = tmp;
    }
    if(r2 - r1 >= d)
        return pi*r1*r1;
    double ang1=acos((r1*r1+d*d-r2*r2)/(2*r1*d));
    double ang2=acos((r2*r2+d*d-r1*r1)/(2*r2*d));
    return ang1*r1*r1 + ang2*r2*r2 - r1*d*sin(ang1);
}
bool check(double xx,double yy,double rr)
{
    int i;
    for(i=0;i<n;i++)
    {
        if(mj(xx,yy,rr,x[i],y[i],r[i])*2<r[i]*r[i]*pi)return 0;
    }
    return 1;
}
double calc(double xx,double yy)//二分找满足条件的最小半径
{
    int i;
    double l=0,r=20000;
    while(r-l>1e-6)
    {
        double mid=(l+r)/2.0;
        if(check(xx,yy,mid))r=mid;
        else l=mid;
    }
    return l;
}
void work()
{
    int i;
    double minn=1e9;
    for(i=0;i<n;i++)
    {
        minn=min(minn,calc(x[i],y[i]));//以这个为圆心
    }
    printf("%.4f\n",minn);
}
int main()
{
    int i,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf%lf",&x[i],&y[i],&r[i]);
        }
        work();
    }
    return 0;
}