Buried memory

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来源：HDU - 3007

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题目描述：

Each person had do something foolish along with his or her growth.But,when he or she did this that time,they could not predict that this thing is a mistake and they will want this thing would rather not happened.
The world king Sconbin is not the exception.One day,Sconbin was sleeping,then swakened by one nightmare.It turned out that his love letters to Dufein were made public in his dream.These foolish letters might ruin his throne.Sconbin decided to destroy the letters by the military exercises’s opportunity.The missile is the best weapon.Considered the execution of the missile,Sconbin chose to use one missile with the minimum destruction.
Sconbin had writen N letters to Dufein, she buried these letters on different places.Sconbin got the places by difficult,he wants to know where is the best place launch the missile,and the smallest radius of the burst area. Let’s help Sconbin to get the award.

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输入格式：

There are many test cases.Each case consists of a positive integer N(N<500,^V,our great king might be a considerate lover) on a line followed by N lines giving the coordinates of N letters.Each coordinates have two numbers,x coordinate and y coordinate.N=0 is the end of the input file.

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输出格式：

For each case,there should be a single line in the output,containing three numbers,the first and second are x and y coordinates of the missile to launch,the third is the smallest radius the missile need to destroy all N letters.All output numbers are rounded to the second digit after the decimal point.

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输入样例：

3
1.00 1.00
2.00 2.00
3.00 3.00
0

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输出样例：

2.00 2.00 1.41

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思路：

典型的最小覆盖圆问题，用求解最小覆盖圆问题的算法解决即可。

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AC代码：

#include <bits/stdc  .h>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;

const int INF = 0x3f3f3f3f;
const int EPS = 1e-8;

void __init__ ()
{
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
}

struct Point
{
    double x;
    double y;
};

int sgn (double x)
{
    if (fabs(x) < EPS) return 0;
    return x > 0 ? 1 : -1;
}

double distance (Point a, Point b)
{
    return hypot(a.x - b.x, a.y - b.y);
}

Point circle_center (Point a, Point b, Point c)
{
    Point center;
    double A1 = 2 * (b.x - a.x), B1 = 2 * (b.y - a.y), C1 = b.x * b.x   b.y * b.y - a.x * a.x - a.y * a.y;
    double A2 = 2 * (c.x - b.x), B2 = 2 * (c.y - b.y), C2 = c.x * c.x   c.y * c.y - b.x * b.x - b.y * b.y;
    center.x = (C1 * B2 - C2 * B1) / (A1 * B2 - A2 * B1);
    center.y = (C1 * A2 - C2 * A1) / (B1 * A2 - A1 * B2);
    return center;
}

void minus_cover_circle (vector<Point> p, Point& c, double& r)
{
    random_shuffle(p.begin(), p.end());
    c = p[0];
    r = 0;
    for (int i = 1; i < p.size(); i  )
    {
        if (sgn(distance(p[i], c) - r) > 0)
        {
            c = p[i];
            r = 0;
            for (int j = 0; j < i; j  )
            {
                if (sgn(distance(p[j], c) - r) > 0)
                {
                    c.x = (p[i].x   p[j].x) / 2;
                    c.y = (p[i].y   p[j].y) / 2;
                    r = distance(c, p[j]);
                    for (int k = 0; k < j; k  )
                    {
                        if (sgn(distance(p[k], c) - r) > 0)
                        {
                            c = circle_center(p[i], p[j], p[k]);
                            r = distance(c, p[k]);
                        }
                    }
                }
            }
        }
    }
}

vector<Point> p;

int main ()
{

    __init__();

    int n;
    Point c;
    double r;
    while (cin >> n && n)
    {
        p.resize(n);
        for (int i = 0; i < n; i  ) cin >> p[i].x >> p[i].y;
        minus_cover_circle(p, c, r);
        cout << fixed << setprecision(2) << c.x << " " << c.y << " " << r << endl;
    }

    return 0;
}

来源：https://www.icode9.com/content-4-891251.html