hdu 2348 Turn the corner(三分&&几何)(中等)

Turn the corner Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2229    Accepted Submission(s): 856

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

题意：

已知汽车的长和宽，l和w。以及俩条路的宽为x和y。汽车所处道路宽为x 。问汽车是否能顺利转弯？

分析：汽车是否能顺利转弯取决于在极限情况下，随着角度的变化，汽车离对面路的距离是否大于等于0

如图中

在上图中须要计算转弯过程中h 的最大值是否小于等于y非常明显。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。能够用三分法来求解

代码：

#include<iostream>
#include<algorithm>
#include<math.h>
#include<cstdio>
using namespace std;
#define pi 3.141592653
double x,y,l,w;
double cal(double a)
{
    double s=l*cos(a)+w*sin(a)-x;
    double h=s*tan(a)+w*cos(a);
    return h;
}
int main()
{
    while(scanf("%lf %lf %lf %lf",&x,&y,&l,&w)!=EOF)
    {
        double left=0.0,right=pi/2;
        double lm,rm;
        while(fabs(right-left)>1e-6)
        {
            lm=(left*2.0+right)/3.0;
            rm=(left+right*2.0)/3.0;
            if(cal(lm)>cal(rm))
                right=rm;
            else left=lm;
        }
        if(cal(left)<=y)
            printf("yes\n");
        else printf("no\n");
    }
    return 0;
}