求某一点到其它点距离和最小，求这个和，这个点 为费马点。

做法：

1 #include 
      
        2 #include
       
         3 #include
        
          4 #include
         
           5 #include
          
            6 #include
           
             7 #include
            
              8 #include
             
               9 #include
              
               10 using namespace std;11 #define N 10000012 #define LL long long13 #define INF 0xfffffff14 const double eps = 1e-8;15 const double pi = acos(-1.0);16 const double inf = ~0u>>2;17 struct point18 {19 double x,y;20 point(double x=0,double y =0):x(x),y(y){}21 }p[N];22 int n;23 int dir[4][2] = { 0,1,0,-1,1,0,-1,0};24 typedef point pointt;25 pointt operator -(point a,point b)26 {27 return point(a.x-b.x,a.y-b.y);28 }29 double dis(point a)30 {31 return sqrt(a.x*a.x+a.y*a.y);32 }33 double cal(point pp)34 {35 int i;36 double s = 0;37 for(i = 1; i <= n; i++)38 s+=dis(pp-p[i]);39 return s;40 }41 int main()42 {43 int i;44 while(scanf("%d",&n)!=EOF)45 {46 double t = 10000;47 point pp = point(0,0);48 for(i =1 ; i <= n;i ++)49 {50 scanf("%lf%lf",&p[i].x,&p[i].y);51 pp.x+=p[i].x;52 pp.y+=p[i].y;53 }54 pp.x /=n;55 pp.y /=n;56 double sum = cal(pp),tmp;57 point tp,ans = pp;58 while(t>0.02)59 {60 int flag = 1;61 while(flag)62 {63 flag = 0;64 for(i =0 ; i< 4 ; i++)65 {66 tp = point(pp.x+dir[i][0]*t,pp.y+dir[i][1]*t);67 tmp = cal(tp);68 if(tmp