hdu 2438Turn the corner（三分）

Turn the corner

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1613    Accepted Submission(s): 599

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

 

解題思路：

關鍵是要找到小車的運動狀態，下面是分析和公式推導；

在小車轉彎過程中，黃線是不斷地變化的，變化規律是先增大再減小。所以抓住這一點，用三分法。先找一個變量，角度sita（就是上圖中用紅色標記的那個角），之後就是一系列的推導，算出黃線的長度。角度的範圍是（0，pi/2）。

當三分找出最長的黃線長度之後，使之與Y做比較，當它小於Y時，就說明能夠通過了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
#ifdef __int64
typedef __int64 LL;
#else
typedef long long LL;
#endif
#define PI 3.1415926
#define esp 1e-8
double x,y,l,d;
double H_value(double sita,double l,double x,double d)//算出黃線的長度
{
    return l*sin(sita)-x*tan(sita)+d/cos(sita);
}
int main()
{
    while(~scanf("%lf%lf%lf%lf",&x,&y,&l,&d))
    {
        double sita_left=0.0;
        double sita_right=PI/2.0;
        double mid,midmid;
        double mid_value,midmid_value;
        while((sita_right-sita_left)>=esp)//找出最長的黃線狀態
        {
            mid=(sita_left+sita_right)/2.0;
            midmid=(mid+sita_right)/2.0;
            mid_value=H_value(mid,l,x,d);
            midmid_value=H_value(midmid,l,x,d);
            if(mid_value>midmid_value)
                sita_right=midmid;
            else
                sita_left=mid;
        }
        if((y-mid_value)>esp)//如果最長的黃線都比Y小的話，那就說明能夠通過
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}