POJ 3831 & HDU 3264 Open-air shopping malls（几何）

题目链接：

POJ:

id=3831" target="_blank">http://poj.org/problem?id=3831

HDU:http://acm.hdu.edu.cn/showproblem.php?pid=3264

Description

The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. 



Unfortunately, the climate has changed little by little and now rainy days seriously affected the operation of open-air shopping malls -- it's obvious that nobody will have a good mood when shopping in the rain. In order to change this situation, the manager
of these open-air shopping malls would like to build a giant umbrella to solve this problem. 



These shopping malls can be considered as different circles. It is guaranteed that these circles will not intersect with each other and no circles will be contained in another one. The giant umbrella is also a circle. Due to some technical reasons, the center
of the umbrella must coincide with the center of a shopping mall. Furthermore, a fine survey shows that for any mall, covering half of its area is enough for people to seek shelter from the rain, so the task is to decide the minimum radius of the giant umbrella
so that for every shopping mall, the umbrella can cover at least half area of the mall.

Input

The input consists of multiple test cases. 



The first line of the input contains one integer T (1 <= T <= 10), which is the number of test cases. For each test case, there is one integer N (1 <= N <= 20) in the first line, representing the number of shopping malls. 



The following N lines each contain three integers X,Y,R, representing that the mall has a shape of a circle with radius R and its center is positioned at (X, Y). X and Y are in the range of [-10000,10000] and R is a positive integer less than 2000.

Output

For each test case, output one line contains a real number rounded to 4 decimal places, representing the minimum radius of the giant umbrella that meets the demands.

Sample Input

1
2
0 0 1
2 0 1

Sample Output

2.0822

Source

field=source&key=Ningbo+2009" style="text-decoration:none">Ningbo 2009

题意： 

给出一些圆，选择当中一个圆的圆心为圆心。然后画一个大圆。要求大圆最少覆盖每一个圆的一半面积。求最小面积。

代码例如以下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;
const double eps = 1e-8;
const double PI = acos(-1.0);

int sgn(double x)
{
    if(fabs(x) < eps) return 0;
    if(x < 0) return - 1;
    else return 1;
}
struct Point
{
    double x, y, r;
    Point() {}
    Point(double _x, double _y)
    {
        x = _x;
        y = _y;
    }
    Point operator -( const Point &b) const
    {
        return Point(x - b. x, y - b. y);
    }
    //叉积
    double operator ^ (const Point &b) const
    {
        return x*b. y - y*b. x;
    }
    //点积
    double operator * (const Point &b) const
    {
        return x*b. x + y*b. y;
    }
    //绕原点旋转角度B（弧度值），后x,y的变化
    void transXY(double B)
    {
        double tx = x,ty = y;
        x = tx* cos(B) - ty*sin(B);
        y = tx* sin(B) + ty*cos(B);
    }
};
Point p[47];

//*两点间距离
double dist( Point a, Point b)
{
    return sqrt((a-b)*(a- b));
}
//两个圆的公共部分面积
double Area_of_overlap(Point c1, double r1, Point c2, double r2)
{
    double d = dist(c1,c2);
    if(r1 + r2 < d + eps) return 0;
    if(d < fabs(r1 - r2) + eps)
    {
        double r = min(r1,r2);
        return PI*r*r;
    }
    double x = (d*d + r1*r1 - r2*r2)/(2*d);
    double t1 = acos(x / r1);
    double t2 = acos((d - x)/r2);
    return r1*r1*t1 + r2*r2*t2 - d*r1*sin(t1);
}

int main()
{
    double x1, y1, r1, x2, y2, r2;
    int t;
    int n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i = 0; i < n; i++)
        {
            scanf("%lf %lf %lf",&p[i].x,&p[i].y,&p[i].r);
        }
        double ans = 999999;
        double l, r, mid;
        for(int i = 0; i < n; i++) //枚举圆心
        {
            l = 0;
            r = 35000.0;//二分
            while(r-l > eps)//能找到
            {
                mid = (l+r)/2.0;
                int flag = 0;
                for(int j = 0; j < n; j++) // 每一个点
                {
                    if(Area_of_overlap(p[i],mid,p[j],p[j].r)<p[j].r*p[j].r*PI/2.0)
                    {
                        flag = 1;//太小
                        break;
                    }
                }
                if(flag)
                    l = mid;
                else
                    r = mid;
            }
            if(l < ans)
                ans = l;
        }
        printf("%.4lf\n",ans);
    }
    return 0;
}