C - Buried memory HDU - 3007
Problem Description
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.
Input
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.
Output
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.
Sample Input
3
1.00 1.00
2.00 2.00
3.00 3.00
0
Sample Output
2.00 2.00 1.41
题意：给几个点，用一个最小的圆把这些点全部覆盖，求这个圆的圆心半径
思路：穷举法，最小圆覆盖必然是某三个顶点的外界圆，那么设置三层循环进行枚举就可以了！
虽然算法是o(n^3),但是实际运行大概是在o(n)左右
但是会出现下面的例外。
也就是说三层遍历得到C(n,3)个外接圆，并不能够保证算法的正确性，我们还要构造C(n,2)个以两个点的连线为直径的圆。
关于求解一个三角形的外接圆圆心：
我们只需找到两条线段的中垂线然后求交点即可。
图片来自于https://blog.csdn.net/migu77777/article/details/52198299
#include <stdio.h>
#include <math.h>
#include <algorithm>
#define eps 1e-8
using namespace std;
struct point
{
double x, y;
}p[510];
point o;
double r;
double dis(point a, point b)
{
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
void getcircle(point a, point b, point c)
{//求外心
double a1 = b.x - a.x, b1 = b.y - a.y, c1 = (a1 * a1 + b1 * b1) / 2;
double a2 = c.x - a.x, b2 = c.y - a.y, c2 = (a2 * a2 + b2 * b2) / 2;
double d = a1 * b2 - a2 * b1;
o.x = a.x + (c1 * b2 - c2 * b1) / d;
o.y = a.y + (a1 * c2 - a2 * c1) / d;
}
int main ()
{
int n;
while(scanf("%d", &n))
{
if(n == 0)
break;
for(int i = 0; i < n; i++)
scanf("%lf %lf", &p[i].x, &p[i].y);
r = 0.0;//半径
o = p[0];//圆心
for(int i = 1; i < n; i++)
{
if(dis(p[i], o) > r )
{//先确定第一个点
r = 0.0;
o = p[i];
for(int j = 0; j < i; j++)
{//确定第二个点
if(dis(p[j], o) > r )
{
o.x = (p[j].x + p[i].x) / 2.0;
o.y = (p[j].y + p[i].y) / 2.0;
r = dis(p[j], o);
for(int k = 0; k < j; k++)
{//如果还有点不在园内，那么就是三个点的情况，确定第三个点
if(dis(p[k], o) > r)
{
getcircle(p[i], p[j], p[k]);
r = dis(p[k], o);
}
}
}
}
}
}
printf("%.2lf %.2lf %.2lf\n", o.x, o.y, r);
}
return 0;
}