详见：
http://blog.csdn.net/popoqqq/article/details/42366599
http://blog.csdn.net/whzzt/article/details/51346228
用拉格朗日乘数法，求了偏导之后二分λ。然后求完偏导的那个一元三次式的解可以二分求，因为是单调递增的。
总复杂度( O(nlog^2n) )
#include<cstdio>
#include<cmath>
using namespace std;
const int N=120005;
int n,q,op,i,m;
double X1,X2,Y1,Y2,a,b,c,d,aa[N],bb[N],cc[N],ab[N],ac[N],bc[N],saa,sbb,scc,sab,sac,sbc,eps=1e-8,ans;
inline bool cmp(double x)
{
return fabs(x)<eps;
}
inline double solve(double a,double b,double c)
{
if(cmp(a))
return c;
double x=-b/(2.0*a);
return a*x*x+b*x+c;
}
int main()
{
scanf("%d",&q);
while(q--)
{
scanf("%d",&op);
if(op==0)
{
scanf("%lf%lf%lf%lf",&X1,&Y1,&X2,&Y2);
if(cmp(X1-X2))
a=1,b=0,c=-X1;
else
a=(Y2-Y1)/(X2-X1),b=-1,c=Y1-a*X1;
d=a*a+b*b;
aa[++n]=a*a/d,bb[n]=b*b/d,cc[n]=c*c/d,ab[n]=a*b/d,ac[n]=a*c/d,bc[n]=b*c/d;
saa+=aa[n],sbb+=bb[n],scc+=cc[n],sab+=ab[n],sac+=ac[n],sbc+=bc[n];
m++;
}
if(op==1)
{
scanf("%d",&i);
saa-=aa[i],sbb-=bb[i],scc-=cc[i],sab-=ab[i],sac-=ac[i],sbc-=bc[i];
m--;
}
if(op==2)
{
if(!m)
{
puts("0.00");
continue;
}
if(cmp(sbb))
a=b=0;
else
a=-sab/sbb,b=-sbc/sbb;
ans=solve(saa+2.0*a*sab+a*a*sbb,2.0*(b*sab+sac+a*b*sbb+a*sbc),b*b*sbb+2.0*b*sbc+scc);
if(cmp(ans))
ans=0;
printf("%.2f
",ans);
}
}
return 0;
}