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

Posted 2022-07-11 ydw--

 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;

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