BZOJ 2508 简单题/BZOJ 3775 点和直线

求下偏导就好了。。。。直接O（1）维护。

#include<iostream>
#include<cstring>
#include<cstdio> 
#include<algorithm>
#include<cmath>
#define eps 1e-7
#define maxn 120050
using namespace std;
double a=0,b=0,c=0,d=0,e=0,f=0;
double x11,y11,x22,y22;
double tab[10];
int n,type,k,cnt=0,tot=0;
struct line
{
    double a,b,c;
    line (double a,double b,double c):a(a),b(b),c(c) {}
    line () {}
}p[maxn];
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&type);
        if (type==0)
        {
            scanf("%lf%lf%lf%lf",&x11,&y11,&x22,&y22);
            if (x11==x22) p[++cnt]=line(1,0,-x11);
            else if (y11==y22) p[++cnt]=line(0,1,-y11);
            else p[++cnt]=line(y22-y11,x11-x22,-x11*y22+x11*y11+x22*y11-x11*y11);
            double a1=p[cnt].a,b1=p[cnt].b,c1=p[cnt].c,dd=a1*a1+b1*b1;
            a+=a1*a1/dd;b+=a1*b1/dd;c+=a1*c1/dd;
            d+=a1*b1/dd;e+=b1*b1/dd;f+=b1*c1/dd;
            tab[1]+=a1*a1/dd;tab[2]+=b1*b1/dd;tab[3]+=c1*c1/dd;
            tab[4]+=2.0*a1*b1/dd;tab[5]+=2.0*a1*c1/dd;tab[6]+=2.0*b1*c1/dd;
            tot++;
        }
        else if (type==1)
        {
            scanf("%d",&k);tot--;
            double a1=p[k].a,b1=p[k].b,c1=p[k].c,dd=a1*a1+b1*b1;
            a-=a1*a1/dd;b-=a1*b1/dd;c-=a1*c1/dd;
            d-=a1*b1/dd;e-=b1*b1/dd;f-=b1*c1/dd;
            tab[1]-=a1*a1/dd;tab[2]-=b1*b1/dd;tab[3]-=c1*c1/dd;
            tab[4]-=2.0*a1*b1/dd;tab[5]-=2.0*a1*c1/dd;tab[6]-=2.0*b1*c1/dd;
        }
        else
        {
            if (tot==1) {printf("0.00\n");continue;}
            double x,y;
            if (a*e==b*d)
            {
                x=0;
                if (fabs(b)>eps) y=-c/b;
                else if (fabs(e)>eps) y=-(f/e);
                else y=0;
                double ans=tab[1]*x*x+tab[2]*y*y+tab[3]+tab[4]*x*y+tab[5]*x+tab[6]*y;
                y=0;
                if (fabs(a)>eps) x=-(c/a);
                else if (fabs(d)>eps) x=-(f/d);
                else x=0;
                ans=min(ans,tab[1]*x*x+tab[2]*y*y+tab[3]+tab[4]*x*y+tab[5]*x+tab[6]*y);
                printf("%.2f\n",ans);continue;
            }
            else {y=(a*f-c*d)/(b*d-a*e);x=(b*f-c*e)/(a*e-b*d);}
            double ans=tab[1]*x*x+tab[2]*y*y+tab[3]+tab[4]*x*y+tab[5]*x+tab[6]*y;
            if (fabs(ans)<1e-3) printf("0.00\n");else printf("%.2f\n",ans);
        }
    }
    return 0;
}