bzoj1857

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三分套三分

电影课上写的orz

gyz大佬直接秒出正解orz

我只负责敲敲qwq

先三分第一条直线上走的长度，在三分第二条直线上的长度

/**************************************************************

    Problem: 1857

    User: syh0313

    Language: C++

    Result: Accepted

    Time:52 ms

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    Memory:1288 kb

****************************************************************/

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

using namespace std;

const double eps=1e-8;

double ax,ay,bx,by,cx,cy,dx,dy,p,q,r;

double calc(double x,double y)

{

    double aa=x/p; aa+=y/q;

    double xxa,yya,xxb,yyb;

    if (sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by))<eps)

    {xxa=ax; yya=ay;}

    else

    {

      xxa=ax+x*(bx-ax)/sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by));

      yya=ay-x*(ay-by)/sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by));

    }

    if (sqrt((cx-dx)*(cx-dx)+(cy-dy)*(cy-dy))<eps)

    {xxb=dx; yyb=dy;}

    else

    {

      xxb=dx-y*(dx-cx)/sqrt((cx-dx)*(cx-dx)+(cy-dy)*(cy-dy));

      yyb=dy-y*(dy-cy)/sqrt((cx-dx)*(cx-dx)+(cy-dy)*(cy-dy));

    }

    aa+=sqrt((xxa-xxb)*(xxa-xxb)+(yya-yyb)*(yya-yyb))/r;

return aa;

}

double work(double x)

{

    double l=0,r=sqrt((cx-dx)*(cx-dx)+(cy-dy)*(cy-dy));

    while (eps<r-l)

    {

        double lmid=l+(r-l)/3.0,rmid=r-(r-l)/3.0;

        double ti1=calc(x,lmid),ti2=calc(x,rmid);

        if (ti1-ti2>eps) l=lmid;else r=rmid;

    }

return calc(x,l);

}

int main()

{

    scanf("%lf%lf%lf%lf",&ax,&ay,&bx,&by);

    scanf("%lf%lf%lf%lf",&cx,&cy,&dx,&dy);

    scanf("%lf%lf%lf",&p,&q,&r);

    double l=0,r=sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by));

    while (eps<r-l)

    {

        double lmid=l+(r-l)/3.0,rmid=r-(r-l)/3.0;

        double ti1=work(lmid),ti2=work(rmid);

        if (ti1-ti2>eps) l=lmid;else r=rmid;

    }

    printf("%.2lf\n",work(l));

return 0;

}