HD 2438 Turn the corner (三分&&计算几何&&经典问题)

Turn the corner

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

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

思路
这是一个生活中比较常见的问题，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 ，问汽车能否顺利转弯？

分析：汽车能否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（即伸出长度小于等于y）

如下图中 满足：

s=l*cos(angle)+w*sin(angle)-x;
h=s*tan(angle)+w*cos(angle);

在转弯的过程中高度h先增大后减小（角度由pi/2 ->0 ）三分时l=0,r=pi/2就好（相当于角度由0 -> pi/2时，h也是先增大后减小）
即满足单峰函数形式




代码示例

#include<bits/stdc++.h>
using namespace std;
#define eps 1e-6
const double pi=acos(-1.0);

double x,y,l,w;

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

int main()
{
	while(scanf("%lf %lf %lf %lf",&x,&y,&l,&w)!=EOF)
	{
		double l=0,r=pi/2;
		double mid,midmid;
		while(r-l>eps)
		{
			mid=(l*2.0+r)/3.0;//采用除3形式
			midmid=(l+2.0*r)/3.0;
			if(filter(mid)>filter(midmid)) r=midmid;//极大值做法
			else l=mid;
		}
		if(filter(l)<=y) printf("yes\n");
		else printf("no\n");
	}
	return 0;
}