hdu Line belt

简介:

Line belt

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 85 Accepted Submission(s): 47
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
 三分搜索:

如图所示必定在AB CD上存在XY 使得A---D 用时间最短。用三分法针对AB间的某个点找出在CD段上最小的点,

嵌套循环找出最小的时间。有关时间函数在AB段和在CD段均为凸函数,所以可以用三分法。


复制代码
#include <iostream>
#include
<stdio.h>
#include
<math.h>
using namespace std;
struct point
{
double x,y;
};
int p,q,r,v;
double dis(point a,point b)
{
return pow(pow(a.y-b.y,2)+pow(a.x-b.x,2),1.0/2);
}
double findy(point c,point d,point y)
{
point mid,midmid,left,right;
double t1,t2;
left
= c;
right
= d;
do
{
mid.x
= (left.x+right.x)/2;
mid.y
= (left.y+right.y)/2;
midmid.x
= (right.x+mid.x)/2;
midmid.y
= (right.y+mid.y)/2;
t1
= dis(d,mid)/q+dis(mid,y)/r;
t2
= dis(d,midmid)/q+dis(midmid,y)/r;
if(t1>t2)
left
= mid;
else right = midmid;
}
while(fabs(t1-t2)>0.000001);
return t1;
}
double find(point a,point b,point c,point d)
{
point mid,midmid,left,right;
double t1,t2;
left
= a;
right
= b;
do
{
mid.x
= (left.x+right.x)/2;
mid.y
= (left.y+right.y)/2;
midmid.x
= (right.x+mid.x)/2;
midmid.y
= (right.y+mid.y)/2;
t1
= dis(a,mid)/p+findy(c,d,mid);
t2
= dis(a,midmid)/p+findy(c,d,midmid);
if(t1>t2)left = mid;
else right = midmid;
}
while(fabs(t1-t2)>0.000001);
return t1;
}
int main()
{
int t;
point a,b,c,d;
scanf(
"%d",&t);
while(t--)
{
scanf(
"%lf%lf%lf%lf %lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y,&c.x,&c.y,&d.x,&d.y);
scanf(
"%d%d%d",&p,&q,&r);
printf(
"%.2f\n",find(a,b,c,d));
}
return 0;
}
复制代码
















本文转自NewPanderKing51CTO博客,原文链接: http://www.cnblogs.com/newpanderking/archive/2011/08/27/2155430.html  ,如需转载请自行联系原作者

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