Turn the corner

最新推荐文章于 2018-03-26 16:19:25 发布

转载最新推荐文章于 2018-03-26 16:19:25 发布 · 105 阅读

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原文链接：http://www.cnblogs.com/wuwangchuxin0924/p/5996442.html

文章标签：

#java

本文介绍了一个判断汽车能否顺利通过垂直拐角的算法。该算法基于输入的街道宽度及汽车尺寸，通过计算确定汽车是否能够完成转弯。使用了二分查找法来逼近最优解。

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Turn the corner

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 654 Accepted Submission(s): 291

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

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gaojie

/*
x和y值和题目相反，搞了了好几遍
*/
#include<bits/stdc++.h>
#define exp 1e-9
#define pi acos(-1)
using namespace std;
/*
小车不靠墙壁段的直线方程 
*/
double dis(double Q,double l,double d,double x)
{
    double b=d/cos(Q)+l*sin(Q);
    return (b-x)/tan(Q);
}
int main()
{
    //freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
    double x,y,l,d;
    while(scanf("%lf%lf%lf%lf",&x,&y,&l,&d)!=EOF)
    {
        double left=0,right=pi/2,mid,midmid;
        //cout<<right<<endl;
        while(right-left>exp)
        {
            mid=left+(right-left)/3;
            midmid=right-(right-left)/3;
            //假设求解最大极值.
            if(dis(mid,l,d,x)>=dis(midmid,l,d,x))
                right=midmid;
            else 
                left=mid;
        }
        //cout<<"mid="<<mid<<endl;
        //cout<<(int)dis(mid,l,d,x)<<endl;
        if(dis(mid,l,d,x)<y)
            puts("yes");
        else
            puts("no");
    }
    return 0;
}