bzoj1857: [Scoi2010]传送带

显然时间函数有单调性，自然会想到三分。（貌似大家都知道这是三分模板题。。。）

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;const double eps=1e-4;int ax,ay,bx,by,cx,cy,dx,dy,p,q,r;double ans=0;double dis(double ax,double ay,double bx,double by){    return sqrt((bx-ax)*(bx-ax)+(by-ay)*(by-ay));}double work(double sx,double sy,double x,double y){    return dis(ax,ay,sx,sy)/p+dis(sx,sy,x,y)/r+dis(x,y,dx,dy)/q;}double calc(double sx,double sy){    double lx=cx,ly=cy,rx=dx,ry=dy;    while(fabs(lx-rx)>eps||fabs(ly-ry)>eps)    {        double mx=lx+(rx-lx)/3,my=ly+(ry-ly)/3,mmx=mx+(rx-mx)/2,mmy=my+(ry-my)/2;        if(work(sx,sy,mx,my)>work(sx,sy,mmx,mmy))        {            lx=mx;ly=my;        }        else         {            rx=mmx;ry=mmy;        }    }    return work(sx,sy,lx,ly);}int main(){    scanf("%d%d%d%d%d%d%d%d%d%d%d",&ax,&ay,&bx,&by,&cx,&cy,&dx,&dy,&p,&q,&r);    double lx=ax,ly=ay,rx=bx,ry=by;    while(fabs(lx-rx)>eps||fabs(ly-ry)>eps)    {        double mx=lx+(rx-lx)/3,my=ly+(ry-ly)/3,mmx=mx+(rx-mx)/2,mmy=my+(ry-my)/2;        if(calc(mx,my)>calc(mmx,mmy))        {            lx=mx;ly=my;        }        else         {            rx=mmx;ry=mmy;        }    }    ans=min(calc(lx,ly),calc(rx,ry));    printf("%.2lf",ans);    return 0;}