hdu-3007(计算几何+最小覆盖圆)

问题描述:

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.

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

题目题意:题目给我们n个点的坐标，让我们求一个最小圆（可以覆盖所有点）的圆心坐标和半径。

题目分析：题目就一个经典问题，最小圆覆盖问题。

学习了大牛的博客：点击打开链接和点击打开链接。

模板可以自己保存下来咯！

代码如下:

#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
using namespace std;
struct Point
{
double x,y;
};
struct Point a[1005],d;
double r;

double get_dis(Point p1,Point p2) //两点间距离
{
return (sqrt((p1.x-p2.x)*(p1.x -p2.x)+(p1.y-p2.y)*(p1.y-p2.y)));
}
double get_muti(Point p1, Point p2,Point p0)
{
return ((p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y));
}
void get_o(Point p,Point q,int n)
{
d.x=(p.x+q.x)/2.0;
d.y=(p.y+q.y)/2.0;
r=get_dis(p,q)/2;
int k;
double c1,c2,t1,t2,t3;
for(k=1;k<=n;k++) {
if(get_dis(d,a[k])<=r)continue;
if(get_muti(p,q,a[k])!=0.0) {
c1=(p.x*p.x+p.y*p.y-q.x*q.x-q.y*q.y)/2.0;
c2=(p.x*p.x+p.y*p.y-a[k].x*a[k].x-a[k].y*a[k].y)/2.0;
d.x=(c1*(p.y-a[k].y)-c2*(p.y-q.y))/((p.x-q.x)*(p.y-a[k].y)-(p.x-a[k].x)*(p.y-q.y));
d.y=(c1*(p.x-a[k].x)-c2*(p.x-q.x))/((p.y-q.y)*(p.x-a[k].x)-(p.y-a[k].y)*(p.x-q.x));
r=get_dis(d,a[k])
4000
;
}
else {
t1=get_dis(p,q);
t2=get_dis(q,a[k]);
t3=get_dis(p,a[k]);
if(t1>=t2&&t1>=t3) {
d.x=(p.x+q.x)/2.0;
d.y=(p.y+q.y)/2.0;r=get_dis(p,q)/2.0;
}
else if(t2>=t1&&t2>=t3) {
d.x=(a[k].x+q.x)/2.0;
d.y=(a[k].y+q.y)/2.0;
r=get_dis(a[k],q)/2.0;
}
else {
d.x=(a[k].x+p.x)/2.0;
d.y=(a[k].y+p.y)/2.0;
r=get_dis(a[k],p)/2.0;
}
}
}
}

void solve(Point pi,int n)
{
d.x=(pi.x+a[1].x)/2.0;
d.y=(pi.y+a[1].y)/2.0;
r=get_dis(pi,a[1])/2.0;
int j;
for(j=2;j<=n;j++){
if(get_dis(d,a[j])<=r)continue;
else
get_o(pi,a[j],j-1);
}
}
int main()
{
int i,n;
while(scanf("%d",&n)&&n){
for(i=1;i<=n;i++){
scanf("%lf %lf",&a[i].x,&a[i].y);
}
if(n==1) { printf("%.2lf %.2lf 0.00\n",a[1].x,a[1].y);continue;}
r=get_dis(a[1],a[2])/2.0;
d.x=(a[1].x+a[2].x)/2.0;
d.y=(a[1].y+a[2].y)/2.0;
for(i=3;i<=n;i++){
if(get_dis(d,a[i])<=r)continue;
else
solve(a[i],i-1);
}
printf("%.2lf %.2lf %.2lf\n",d.x,d.y,r);
}
return 0;
}