hdu 3400 Line belt（三分）

Line belt

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1865    Accepted Submission(s): 704

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

10 0 0 100100 0 100 1002 2 1

 

Sample Output

136.60

 

Author

lxhgww&&momodi

 

Source

HDOJ Monthly Contest – 2010.05.01

题目大意：

给出两条传送带的起点到末端的坐标，其中ab为p的速度，cd为q的速度 其他地方为r的速度

求a到d点的最短时间

分析：

典型的三分法，先三分第一条线段，找到一个点，然后根据这个点再三分第二条线段即可

ab传送带要有一个间断点mid1， a<mid1<b（从此点跳出ab这条线）同理cd有个间断mid2，路线是a-->mid1-->mid2-->d;

双向三分过程：

默认 mid1=b

   1.由ab对cd取三分，得到mid2点，返回时间t1

            具体为：a到mid1，mid1到cd之间的L点和R点的最小值点mid（做三分），mid到d点，得到三分过 程的最小t1 将mid点赋值给mid2点

   2.由cd对ab求三分确定mid1点，返回时间t2

             由于此时的d到mid2的时间固定，此时只要三分规划出a到mid，mid到mid2的最小值，将mid的值付给mid1

3.重复1和2直到|t1-t2|无穷小为止

我的代码：

#include<iostream>#include<cstdio>#include<cmath>#define eps 1e-15using namespace std;struct point{    double x,y;};point A,B,C,D;double P,Q,R;int sgn(double a){    return (a>eps)-(a<-eps);}double dist(point a,point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}double calcu_lin(point a,point b,point c,point d){    return dist(a,b)/P+dist(b,c)/R+dist(c,d)/Q ;}point MID(point a,point b){    point c;    c.x=(a.x+b.x)*0.5;    c.y=(a.y+b.y)*0.5;    return c;}double sanfen1(point a,point b,point c,point d,point &mid){   point mid1,midmid;    point l,rr;l=c;rr=d;    double t1,t2;   do   {     mid1=MID(l,rr);     midmid=MID(mid1,rr);     t1=calcu_lin(a,b,mid1,d);     t2=calcu_lin(a,b,midmid,d);     if(t1<t2)     rr=midmid;     else     l=mid1;   }   while(sgn(t1-t2)!=0);   mid=mid1;   return  t1;}double sanfen2(point d,point c,point b,point a,point &mid){   point mid1,midmid;    point l,rr;l=a;rr=b;    double t1,t2;   do   {     mid1=MID(l,rr);     midmid=MID(mid1,rr);     t1=calcu_lin(a,mid1,c,d);     t2=calcu_lin(a,midmid,c,d);     if(t1<t2)     rr=midmid;     else     l=mid1;   }   while(sgn(t1-t2)!=0);   mid=mid1;   return  t1;}void work(){    double t1,t2;point mid1,mid2;    mid1.x=B.x;mid1.y=B.y;    do    {        //printf("okok\n");      t1=sanfen1(A,mid1,C,D,mid2);      t2=sanfen2(D,mid2,B,A,mid1);    }    while(sgn(t1-t2)!=0);    printf("%.2lf\n",t1);}int main(){    int t;    double ff;    scanf("%d",&t);     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);          work();    }    return 0;}