题意：
求一个四边形的费马点。
分析：
模拟退火要么超时要么wa，这题的数据就是不想让随机算法过的。。其实四边形的费马点很简单，如果是凸四边形的话费马点是对角线交点，如果是凹四边形费马点是凹点。但题目给的四个点顺序是不确定的，所以要先求下凸包。
代码：
//poj 3990
//sep9
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;struct Point
{double x,y,v;Point(){}Point(double x,double y):x(x),y(y){}
}pnt[8],rec[8];double getSum(Point p)
{double sum=0;for(int i=0;i<4;++i)sum+=sqrt((p.x-pnt[i].x)*(p.x-pnt[i].x)+(p.y-pnt[i].y)*(p.y-pnt[i].y));return sum;
}int cmp(Point a,Point b)
{if(a.y!=b.y) return a.y<b.y;return a.x<b.x;
}int dbl(double x)
{if(fabs(x)<1e-7)return 0;return x>0?1:-1;
}double dis(Point a,Point b)
{return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}int cross(Point a,Point b,Point c)
{double x1=b.x-a.x;double y1=b.y-a.y;double x2=c.x-a.x;double y2=c.y-a.y;return dbl(x1*y2-x2*y1);
}int graham()
{int n=4;sort(pnt,pnt+n,cmp);rec[0]=pnt[0]; rec[1]=pnt[1];int i,pos=1;for(i=2;i<n;++i){while(pos>0&&cross(rec[pos-1],rec[pos],pnt[i])<=0) --pos;rec[++pos]=pnt[i]; }int bak=pos;for(i=n-1;i>=0;--i){while(pos>bak&&cross(rec[pos-1],rec[pos],pnt[i])<=0) --pos;rec[++pos]=pnt[i];}return pos;
}int main()
{int i;while(1){for(i=0;i<4;++i)scanf("%lf%lf",&pnt[i].x,&pnt[i].y);double ans=0;if(pnt[0].x<-0.5) break;if(graham()==4){ans+=dis(rec[0],rec[2]);ans+=dis(rec[1],rec[3]);} else{ans=1e10;for(int i=0;i<4;++i)ans=min(ans,getSum(pnt[i]));}printf("%.4lf\n",ans+1e-8);}return 0;
}