HDU 2438 Turn the corner(三分查找)

托一个学弟的福，学了一下他的最简便三分写法，然后找了一道三分的题验证了下，AC了一题，写法确实方便，还是我太弱了，漫漫AC路！各路大神，以后你们有啥好的简便写法可以在博客下方留个言或私信我，谢谢了！

Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3196 Accepted Submission(s): 1302

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2438

题意：

给出街道在x轴的宽度X，y轴的宽度Y，还有车的长l和宽w，判断是否能够转弯成功。

题解：

盗网上大牛一张图，画的很详细

尽可能让车贴着外面的墙璧转弯，也就是图中的x轴和y轴，此时红线的方程就是图中的方程，此时p点的位置就是让y=X时解得的x值，要保证p点在Y内，也就是-x<y,假若在转弯的所有角度中都满足这个条件，那么就能转弯，分析得，-x先增大后减小，所以用三分求最大-x值。

下面给出AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 double x,y,l,w;
 4 const double eps=1e-6;
 5 const double pi=acos(-1.0);
 6 double solve(double angle)
 7 {
 8     return (-x+l*sin(angle)+w/cos(angle))/tan(angle);
 9 }
10 int main()
11 {
12     while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)!=EOF)
13     {
14         double ll=0,rr=pi/2,midx,midy;
15         while(rr-ll>eps)
16         {
17             midx=(ll+ll+rr)/3;
18             midy=(ll+rr+rr)/3;
19             if(solve(midx)>solve(midy))
20             rr=midy;
21             else ll=midx;
22         }
23         if(solve(ll)<y)
24             printf("yes\n");
25         else printf("no\n");
26     }
27     return 0;
28 }