HDU 2438 Turn the corner(三分枚举角度)

题目大意：给你一个拐角，两边的路的宽度分别为x，y。汽车的长和宽分别为h，w。问你这个汽车否转弯成功。

解题思路：如图，枚举角度。



这是一个凸函数所以三分枚举角度。

Turn the corner

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

Total Submission(s): 2009    Accepted Submission(s): 765


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

 

#include <iostream>
#include<time.h>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<cmath>
#include<map>

#define eps 1e-10
#define PI acos(-1)

using namespace std;

const int maxn = 220;

#define LL long long

double x, y, l, w;

double Del(double ang)
{
double s = l*cos(ang)+w*sin(ang)-x;
double h = s*tan(ang)+w*cos(ang);
return h;
}

int main()
{
while(cin >>x>>y>>l>>w)
{

double Max = 0.0;
double left = 0;
double right = PI/2.0;
while(right-left > eps)
{
double mid = (right+left)/2.0;
double rmid = (mid+right)/2.0;
double xp1 = Del(mid);
double xp2 = Del(rmid);
if(xp1 > xp2) right = rmid;
else left = mid;
Max = max(Max, max(xp1, xp2));
}
if(Max > y) cout<<"no"<<endl;
else cout<<"yes"<<endl;
}
}