Buried memory

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.

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.

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.

31.00 1.002.00 2.003.00 3.000

2.00 2.00 1.41

解题：

     

1.我们先照第一图画出两个圆，画完可知其余的点一定在两圆相交的椭圆里，因为距离就红线最长了，但是范围还是太大了； 2.再照第二个图画出第三个圆，我们可知，    最完美的情况下，最长的线段即是要求的圆的直径，可以包含了所有的其余点，那么这些点就在红色的圆里面；    非完美情况下，即有几个点在红圆的外面（不超出椭圆），那么红圆圆心到那些点最长的，定是所要求的圆的半径。画出第三个图。 (经 提醒，少考虑了等边三角形情况，现补充如下)

3.若是等边三角形，则圆心为等边三角形外心

本来都做对了，奈何一个函数名与系统的库函数冲突了，就是distance函数，CE了3次。

#include <iostream>#include <iomanip>#include <cmath>using namespace std;#define eps 1e-8#define MaxSize 500struct point{ double x,y;};double distance2(point p1,point p2){ return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));}point maxDist(point aPoint[],int n,double& max1Dist){ int i,j,a,b; point bPoint; for (i=0;i<n;i++) { for (j=i+1;j<n;j++) { if (max1Dist<distance2(aPoint[i],aPoint[j])) { max1Dist=distance2(aPoint[i],aPoint[j]); a=i,b=j; } } } bPoint.x=(aPoint[a].x+aPoint[b].x)/2; bPoint.y=(aPoint[a].y+aPoint[b].y)/2; return bPoint;}void maxDist2(point& bPoint,point aPoint[],int n,double& max1Dist){ int i,c; double max2Dist=max1Dist/2; for (i=0;i<n;i++) { if (max2Dist<distance2(bPoint,aPoint[i])) { max2Dist=distance2(bPoint,aPoint[i]); c=i; } } if ((int)(max2Dist*1000)<=(int)(sqrt(max1Dist*max1Dist*3/4)*1000) && (int)((max2Dist+0.01)*1000)>(int)(sqrt(max1Dist*max1Dist*3/4)*1000)) { max1Dist=distance2(aPoint[c],bPoint)*2/3; bPoint.x=(aPoint[c].x+bPoint.x)/3; bPoint.y=(aPoint[c].y+bPoint.y)/3; } else if (max2Dist*2>max1Dist) { max1Dist=max2Dist; } else { max1Dist/=2; }}int main(){ int n,i; double radius; point missile; point everyPoint[MaxSize]; while (cin>>n && n!=0) { radius=0.0; for (i=0;i<n;i++) { cin>>everyPoint[i].x>>everyPoint[i].y; } missile=maxDist(everyPoint,n,radius); maxDist2(missile,everyPoint,n,radius); cout<<fixed<<setprecision(2)<<missile.x<<" "<<missile.y<<" "<<radius<<endl; } return 0;}