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：

#include <bits/stdc++.h>
using namespace std;
double x,y,l,w;
const double eps=1e-6;
const double pi=acos(-1.0);
double solve(double angle)
{
    return (-x+l*sin(angle)+w/cos(angle))/tan(angle);
}
int main()
{
    while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)!=EOF)
    {
        double ll=0,rr=pi/2,midx,midy;
        while(rr-ll>eps)
        {
            midx=(ll+ll+rr)/3;
            midy=(ll+rr+rr)/3;
            if(solve(midx)>solve(midy))
            rr=midy;
            else ll=midx;
        }
        if(solve(ll)<y)
            printf("yes\n");
        else printf("no\n");
    }
    return 0;
}