HDU 3400（搜索题，三分~~初识三分查找）

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 698    Accepted Submission(s): 243

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    先对AB段三分一次，在对CD三分一次。。。一直循环
   
   
   
   
   
   
   
   

    
    
    #include <iostream>
#include <math.h>
using namespace std;
#define EPS 1e-10

struct point 
{
    double x,y;
};

point A,B,C,D,aa,bb,low1,high1,low2,high2,mid1,mid2,mid3,mid4;
double P,Q,R;

double lenth(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

bool judge1(point x1,point x2)
{
    if((lenth(A,x1)/P+lenth(x1,bb)/R+lenth(bb,D)/Q)>=(lenth(A,x2)/P+lenth(x2,bb)/R+lenth(bb,D)/Q))
        return true;
    else return false;
}

bool judge2(point x1,point x2)
{
    if((lenth(A,aa)/P+lenth(aa,x1)/R+lenth(x1,D)/Q)>=(lenth(A,aa)/P+lenth(aa,x2)/R+lenth(x2,D)/Q))
        return true;
    else return false;
}

void san_fen1();
void san_fen2();

int main()
{
    int t;
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            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);
            aa=A;
            bb=D;
            low1=A;
            high1=B;
            low2=C;
            high2=D;
            if(R>=P && R>=Q)
            {
                printf("%.2lf/n",lenth(A,D)/R);
                continue;
            }
            san_fen1();
            printf("%.2lf/n",lenth(A,low1)/P+lenth(low1,low2)/R+lenth(low2,D)/Q);
        }
    }
    return 0;
}

void san_fen1()
{
    while (lenth(low1,high1)>EPS)
    {
        mid1.x=(2*low1.x+high1.x)/3;
        mid1.y=(2*low1.y+high1.y)/3;
        mid2.x=(low1.x+2*high1.x)/3;
        mid2.y=(low1.y+2*high1.y)/3;
        if(judge1(mid1,mid2))
            low1=mid1;
        else
            high1=mid2;
        aa.x=(low1.x+high1.x)/2;
        aa.y=(low1.y+high1.y)/2;
        san_fen2();
    }
    while (lenth(low2,high2)>EPS)
    {
        mid3.x=(2*low2.x+high2.x)/3;
        mid3.y=(2*low2.y+high2.y)/3;
        mid4.x=(low2.x+2*high2.x)/3;
        mid4.y=(low2.y+2*high2.y)/3;
        if(judge2(mid3,mid4))
            low2=mid3;
        else
            high2=mid4;
        bb.x=(low2.x+high2.x)/2;
        bb.y=(low2.y+high2.y)/2;
    }
}

void san_fen2()
{
    if (lenth(low2,high2)>EPS)
    {
        mid3.x=(2*low2.x+high2.x)/3;
        mid3.y=(2*low2.y+high2.y)/3;
        mid4.x=(low2.x+2*high2.x)/3;
        mid4.y=(low2.y+2*high2.y)/3;
        if(judge2(mid3,mid4))
            low2=mid3;
        else
            high2=mid4;
        bb.x=(low2.x+high2.x)/2;
        bb.y=(low2.y+high2.y)/2;
    }
}