HDOJ 3007 Buried memory 增量法最小圓覆蓋

標籤：

增量法最小圓覆蓋,簡單模版

Buried memoryTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2541    Accepted Submission(s): 1365


Problem DescriptionEach 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. 
InputThere 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. 
OutputFor 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

31.00 1.002.00 2.003.00 3.000

 
Sample Output

2.00 2.00 1.41

 
Source2009 Multi-University Training Contest 11 - Host by HRBEU 

/* ***********************************************Author        :CKbossCreated Time  :2014年12月29日 星期一 15時15分46秒File Name     :HDOJ3007.cpp************************************************ */#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <string>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>using namespace std;const int maxn = 888;const double eps=1e-8;struct Point{double x,y;}pt[maxn];Point operator+(Point A,Point B) { return (Point){A.x+B.x,A.y+B.y}; }Point operator-(Point A,Point B) { return (Point){A.x-B.x,A.y-B.y}; }Point operator*(Point A,double p) { return (Point){A.x*p,A.y*p}; }Point operator/(Point A,double p) { return (Point){A.x/p,A.y/p}; }int n;int dcmp(double x){if(fabs(x)<eps) return 0;if(x>eps) return 1;else return -1;}double Cross(Point A,Point B) { return A.x*B.y-A.y*B.x; }double Dot(Point a,Point b) { return a.x*b.x+a.y*b.y; }double Length(Point a) { return sqrt(Dot(a,a)); }struct Circle{Point c; double r;};Circle CircumscribedCircle(Point p1,Point p2,Point p3){double Bx=p2.x-p1.x,By=p2.y-p1.y;double Cx=p3.x-p1.x,Cy=p3.y-p1.y;double D=2*(Bx*Cy-By*Cx);double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x;double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y;Circle c;c.c=(Point){cx,cy};c.r=Length(p1-c.c);return c;}void min_cover_circle(Point p[],int n,Circle& c){c.c=p[0]; c.r=0;for(int i=1;i<n;i++){if(dcmp(Length(p[i]-c.c)-c.r)>0){c.c=p[i]; c.r=0;for(int j=0;j<i;j++){if(dcmp(Length(p[j]-c.c)-c.r)>0){c.c=(Point){(p[i].x+p[j].x)/2,(p[i].y+p[j].y)/2};c.r=Length(p[j]-p[i])/2.;for(int k=0;k<j;k++){if(dcmp(Length(p[k]-c.c)-c.r)>0){c=CircumscribedCircle(p[i],p[j],p[k]);}}}}}}}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);while(scanf("%d",&n)!=EOF&&n){for(int i=0;i<n;i++) scanf("%lf%lf",&pt[i].x,&pt[i].y);Circle c;min_cover_circle(pt,n,c);printf("%.2lf %.2lf %.2lf\n",c.c.x,c.c.y,c.r);}        return 0;}

HDOJ 3007 Buried memory 增量法最小圓覆蓋