Line belt+hdu+三分搜索

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2782    Accepted Submission(s): 1062

Problem Description

In a two-dimensional plane there are two line belts, there are two segments AB and CD, lxhgww's speed on AB is P and on CD is Q, he can move with the speed R on other area on the plane.
How long must he take to travel from A to D?

 

Input

The first line is the case number T.
For each case, there are three lines.
The first line, four integers, the coordinates of A and B: Ax Ay Bx By.
The second line , four integers, the coordinates of C and D:Cx Cy Dx Dy.
The third line, three integers, P Q R.
0<= Ax，Ay，Bx，By，Cx，Cy，Dx，Dy<=1000
1<=P，Q，R<=10

 

Output

The minimum time to travel from A to D, round to two decimals.

 

Sample Input

1 0 0 0 100 100 0 100 100 2 2 1

 

Sample Output

136.60

解决方案：先对AB三分，再对CD三分，AB三分在外层，CD三分在内层。求距离是要注意先加个eps再开放，防止开方结果比真实值小。

code：

#include 
#include
#include
using namespace std;
double dist(double ax,double ay,double bx,double by)
{
    return sqrt((ax-bx)*(ax-bx)+(ay-by)*(ay-by)+0.0000001);
}
double Ax,Ay,Bx,By,Cx,Cy,Dx,Dy;
double P,Q,R;
double AB,CD;
double ax;
double by;
double cal2(double mid)
{

    double cx=(Dx-Cx)*(mid/CD)+Cx;
    double dy=(Dy-Cy)*(mid/CD)+Cy;
    double sum1=dist(ax,by,cx,dy);
    sum1=sum1/R;
    sum1+=(CD-mid)/Q;
    return sum1;
}
double cal(double Mid)
{
    double sum=Mid/P;
    ax=(Bx-Ax)*(Mid/AB)+Ax;
    by=(By-Ay)*(Mid/AB)+Ay;
    CD=dist(Cx,Cy,Dx,Dy);
    double left=0.0,right=CD,mid,midmid,midv,midmidv;
    while(fabs(right-left)>0.000000001)
    {
        mid=(right+left)/2.0;
        midmid=(right+mid)/2.0;
        midv=cal2(mid);
        midmidv=cal2(midmid);
        if(midv<=midmidv)
        {
            right=midmid;
        }
        else left=mid;
    }
    return sum+cal2(left);
}
int main()
{
    int t;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &Ax,&Ay,&Bx,&By,&Cx,&Cy,&Dx,&Dy);
        scanf("%lf%lf%lf",&P,&Q,&R);
        AB=dist(Ax,Ay,Bx,By);
        double left=0.0,right=AB,mid,midmid,midv,midmidv;
        while(fabs(right-left)>0.000000001)
        {
            mid=(right+left)/2.0;
            midmid=(mid+right)/2.0;
            midv=cal(mid);
            midmidv=cal(midmid);
            if(midv<=midmidv)
            {
                right=midmid;
            }
            else left=mid;
        }
        double s=cal(left);
        printf("%.2lf\n",s);
    }
    return 0;
}