heru 5081 Turn the corner（三分）

Turn the corner

TimeLimit: 1 Second MemoryLimit: 32 Megabyte

Totalsubmit: 637 Accepted: 175

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 

heru 5081 Turn the corner 08年哈尔滨regional网络赛
http://acm.hrbeu.edu.cn/index.php?act=problem&id=1280

出处：http://hi.baidu.com/vfxupdpaipbcpuq/item/81b21d1910ea729c99ce33db

汽车拐弯问题，给定X, Y, l, d判断是否能够拐弯。首先当X或者Y小于d，那么一定不能。
其次我们发现随着角度θ的增大，最大高度ｈ先增长后减小，即为凸性函数，可以用三分法来求解。

这里的Calc函数需要比较繁琐的推倒公式：
s = l * cos(θ) + w * sin(θ) - x;
h = s * tan(θ) + w * cos(θ);
其中s为汽车最右边的点离拐角的水平距离, h为里拐点最高的距离, θ范围从0到90。

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
#include<math.h>
#define eps 1e-15
#define PI 3.1415926535897932384626
using namespace std;
double x,y,l,w;
int sgn(double a)
{
    return (a>eps)-(a<-eps);
}
double dist(double a)
{
    double s=l*cos(a)+w*sin(a)-x;
    double h=s*tan(a)+w*cos(a);
    return h;
}
double work()
{
    double mid,midmid;
    double l,r;l=0;r=PI*0.5;
   while(sgn(r-l)>0)
   {
     mid=(l+r)*0.5;
     midmid=(mid+r)*0.5;
     if(dist(mid)>=dist(midmid))
     r=midmid;
     else
     l=mid;
   }
   return dist(r);
}
int main()
{
    double ff;
     while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w))
    {
        if(w>x||w>y)
        printf("yes\n");
        else
        {
            ff=work();
        if(sgn(y-ff)>0)
        printf("yes\n");
        else
        printf("no\n");
        }
    }

    return 0;
}