hdu 2438 Turn the corner [ 三分 ]

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Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2081    Accepted Submission(s): 787

Problem Description

Mr. West bought a new car! So he is travelling around the city.

One
day he comes to a vertical corner. The street he is currently in has a
width x, the street he wants to turn to has a width y. The car has a
length l and a width d.

Can Mr. West go across the corner?

 

Input

Every line has four real numbers, x, y, l and w.
Proceed to the end of file.

 

Output

If he can go across the corner, print "yes". Print "no" otherwise.

 

Sample Input

10 6 13.5 4
10 6 14.5 4

 

Sample Output

yes
no

 

Source

2008 Asia Harbin Regional Contest Online

 

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题解转一发：

http://www.cnblogs.com/wally/archive/2013/05/14/3077383.html

 

网上大牛思路：

可以根据边界，汽车已经转弯，设水平方向为x轴，垂直方向为y轴。

则汽车的内边界(靠近里面的边界)的直线方程式f(x)为：y=x*tan(a)+l*sin(a)+d/cos(a).其中a是汽车与x轴的夹角

当y=X时，求解出的-x即为汽车的内边界到y轴的距离h，若h小于Y即可转弯，若大于Y就不能转弯。
所以只需要利用方程式，求-x的最大值，即可判断能否通过。
由于f(x)是凸函数(随着x的增大y先增大后减小)，所以，需要借助三分求解。

图示：

第一道三分求极值题啊！！！

Run ID	Submit Time	Judge Status	Pro.ID	Exe.Time	Exe.Memory	Code Len.	Language	Author
13216500	2015-03-23 09:01:02	Accepted	2438	0MS	1688K	946 B	G++	czy

 #include <cstdio>
 #include <cmath>

 using namespace std;

 const double mi = 1e-;
 const double eps = 1e-;
 const double pi = 4.0 * atan(1.0);

 double x,y,l,d;

 double cal(double a)
 {
     return (-x + l * sin(a) + d / cos(a)) / tan(a) ;
 }

 int main()
 {
     while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF)
     {
         if(d>y || d>x){
             printf("no\n");continue;
         }
         double low=,high = pi / ,mid,mmid;
         double te1,te2;
         while(high - low > mi)
         {
             mid = (low + high) / ;
             mmid = (low + mid) / ;
             te1 = cal(mid);
             te2 = cal(mmid);
             if(te1 > te2){
                 low = mmid;
             }
             else{
                 high = mid;
             }
         }
         te1 = cal(low);
         if(te1 < y){
             printf("yes\n");
         }
         else{
             printf("no\n");
         }
     }
 }