[BZOJ 1857][SCOI 2010]传送带(三分套三分)

题目链接

http://www.lydsy.com/JudgeOnline/problem.php?id=1857

思路

三分套三分，首先三分在传送带AB上的运动时间，然后在确定了在AB上的运动时间后，三分在CD上的运动时间。
如果此题没有单峰函数的特征，我想也只能用退火套退火乱搞了。。。
当然我刚开始比较naive，真的就去三分运动时间了，这样做比较麻烦，后来翻了下hzwer的题解发现了一个很不错的做法：不三分运动时间，而是三分离开这条传送带时的坐标，这样做就简单多了。

代码

蒟蒻太傻逼，刚开始弄成了求最大值，一直调不出来，然后三分还有几个地方写挂了QAQ

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <cmath>
#define EPS 1e-4

using namespace std;

double P,Q,R,ax,ay,bx,by,cx,cy,dx,dy;

inline double dist(double x1,double y1,double x2,double y2)
{
    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

inline double calc(double sx,double sy) //在(sx,sy)处离开了传送带AB
{
    double lowerBoundx=cx,upperBoundx=dx;
    double lowerBoundy=cy,upperBoundy=dy;
    while(fabs(upperBoundx-lowerBoundx)>EPS||fabs(upperBoundy-lowerBoundy)>EPS)
    {
        double midx1=lowerBoundx+(upperBoundx-lowerBoundx)/3,midx2=upperBoundx-(upperBoundx-lowerBoundx)/3;
        double midy1=lowerBoundy+(upperBoundy-lowerBoundy)/3,midy2=upperBoundy-(upperBoundy-lowerBoundy)/3;
        double f1=dist(ax,ay,sx,sy)/P+dist(sx,sy,midx1,midy1)/R+dist(midx1,midy1,dx,dy)/Q;
        double f2=dist(ax,ay,sx,sy)/P+dist(sx,sy,midx2,midy2)/R+dist(midx2,midy2,dx,dy)/Q;
        if(f1<f2) upperBoundx=midx2,upperBoundy=midy2; //!!!!!
        else lowerBoundx=midx1,lowerBoundy=midy1; //!!!!!
    }
    return dist(ax,ay,sx,sy)/P+dist(sx,sy,lowerBoundx,lowerBoundy)/R+dist(lowerBoundx,lowerBoundy,dx,dy)/Q;
}

int main()
{
    scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf",&ax,&ay,&bx,&by,&cx,&cy,&dx,&dy,&P,&Q,&R);
    double lowerBoundx=ax,upperBoundx=bx;
    double lowerBoundy=ay,upperBoundy=by;
    while(fabs(upperBoundx-lowerBoundx)>EPS||fabs(upperBoundy-lowerBoundy)>EPS)
    {
        double midx1=lowerBoundx+(upperBoundx-lowerBoundx)/3,midx2=upperBoundx-(upperBoundx-lowerBoundx)/3;
        double midy1=lowerBoundy+(upperBoundy-lowerBoundy)/3,midy2=upperBoundy-(upperBoundy-lowerBoundy)/3;
        double f1=calc(midx1,midy1),f2=calc(midx2,midy2);
        if(f1<f2) upperBoundx=midx2,upperBoundy=midy2; //!!!!!
        else lowerBoundx=midx1,lowerBoundy=midy1; //!!!!!
    }
    printf("%.2lf",calc(lowerBoundx,lowerBoundy));
    return 0;
}