Turn the corner+hdu+三分搜索

最新推荐文章于 2021-12-01 20:53:09 发布

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最新推荐文章于 2021-12-01 20:53:09 发布

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分类专栏：二分三分枚举+排序

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本文探讨了在给定宽度、长度和角度的情况下，车辆如何顺利通过垂直拐角的问题，并提供了解决方案。

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Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1788 Accepted Submission(s): 672

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
解决方案：汽车转弯时可看到此图：

汽车要想能过这个弯，最好这样转弯，左侧贴近弯道内侧，右下角贴近弯道内侧，这回形成一个a角，然后转弯的过程a角从0到90过程变化，然后就看p点的位置了，若其能过得了这个弯道，其横坐标必须小于或等于y，而p点的横坐标随a点的变化是一个凸函数，所以用上三分搜索。
用上高中知识推得公式：
f(a)=l*cos(a)+(w-x*cos(a))/sin(a);
不过要有个特判，若x<w或y<w也不能通过。
code:#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
double x,y,l,w;
double cal(double mid)
{
    return l*cos(mid)+(w-x*cos(mid))/sin(mid);
}
int main()
{
    while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&w))
    {
        double left=0.0,right=acos(-1.0)/2,mid,midmid,midv,midmidv;
        while(fabs(right-left)>1e-8)
        {
            mid=(right+left)/2.0;
            midmid=(right+mid)/2.0;
            midv=cal(mid);
            midmidv=cal(midmid);
            if(midv>=midmidv)
            {
                right=midmid;
            }
            else left=mid;
        }
        if(cal(mid)>y||w>x||w>y)
        {
            printf("no\n");
        }
        else printf("yes\n");
    }
    return 0;
}