題目描述

傳送門

題解

感覺一下好像在傳送帶上走的太多或者走得太少時間都不是最優的
實際上答案對走的長短單峰
然後就三分套三分…
注意：
①find中的ans一定要有初始值，防止根本不進行三分直接退出
②檢查分母不能為0

程式碼

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

const double eps=1e-9;
int dcmp(double
 
 x)
{
    if (x<=eps&&x>=-eps) return 0;
    return (x>0)?1:-1;
}
struct Vector
{
    double x,y;
    Vector(double X=0,double Y=0)
    {
        x=X,y=Y;
    }
};
typedef Vector Point;
Vector operator + (Vector a,Vector b) {return Vector(a.x+b.x,a.y+b.y);}
Vector operator - (Vector a,Vector b) {return
 
 Vector(a.x-b.x,a.y-b.y);}
Vector operator * (Vector a,double b) {return Vector(a.x*b,a.y*b);}

Point a,b,c,d;
double p,q,r,ans;

double Dot(Vector a,Vector b)
{
    return a.x*b.x+a.y*b.y;
}
double Len(Vector a)
{
    return sqrt(Dot(a,a));
}
double calc(double dis1,double dis2)
{
    Vector v,w;
    Point e,f;
    double
 
 dis3,t;

    v=b-a,w=c-d;
    if (dcmp(Len(v))) v=v*(dis1/Len(v));e=a+v;
    if (dcmp(Len(w))) w=w*(dis2/Len(w));f=d+w;
    dis3=Len(f-e);
    t=dis1/p+dis2/q+dis3/r;
    return t;
}
double find2(double dis)
{
    double l=0,r=Len(d-c),mid1,mid2,ans1,ans2,ans=0;
    while (dcmp(r-l)>0)
    {
        mid1=l+(r-l)/3.0;
        mid2=r-(r-l)/3.0;
        ans1=calc(dis,mid1);
        ans2=calc(dis,mid2);
        if (dcmp(ans1-ans2)<=0) ans=mid1,r=mid2;
        else l=mid1;
    }
    return calc(dis,ans);
}
double find1()
{
    double l=0,r=Len(b-a),mid1,mid2,ans1,ans2,ans=0;
    while (dcmp(r-l)>0)
    {
        mid1=l+(r-l)/3.0;
        mid2=r-(r-l)/3.0;
        ans1=find2(mid1);
        ans2=find2(mid2);
        if (dcmp(ans1-ans2)<=0) ans=mid1,r=mid2;
        else l=mid1;
    }
    return find2(ans);
}
int main()
{
    scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
    scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
    scanf("%lf%lf%lf",&p,&q,&r);
    ans=find1();
    printf("%.2lf\n",ans);
}