hdu 2438 Turn the corner（几何+三分）

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

  13.5
  14.5 

 

Sample Output

yes
no

 

Source

2008 Asia Harbin Regional Contest Online

 

摘自大牛的解题报告:

题意：给定一个直角弯道的两条道路的宽度，然后再给出汽车的长度与宽度，问汽车能否通过该弯道？

 

如下图：

 

要使汽车能转过此弯道，那么就是汽车的左边尽量贴着那个直角点，而汽车的右下后方的点尽量贴着最下面的边。

我们以O点为原点建立直角坐标系，我们可以根据角a给出P点横坐标的函数F(a)

 

那么很容易得到：

 

其中有条件：，可以很容易证明是一个单峰函数，所以接下来就是三分了，如果的最大值小于等于

 

y，那么就能通过此直角弯道，否则就通不过。

 

在三分的时候，写成这样

mid1=(low+high)/;
mid2=(mid1+high)/;

一直WA，直到写成下面这样才AC，

double mid1=(2*low+high)/3; 
double mid2=(low+2*high)/3;

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 int dirx[]={,,-,};
 int diry[]={-,,,};
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1000000
 #define inf 1<<26

 double x,y,l,w;
 double equ(double t)
 {
     return l*cos(t)+(w-x*cos(t))/sin(t);
 }
 int main()
 {

     while(scanf("%lf%lf%lf%lf",&x,&y,&l,&w)==){
         double low=;
         double high=PI/2.0;
     //    double mid1=(low+high)/2;
         //double mid2;
         double sum1;
         double sum2;
         while(high-low>eps){
             double mid1=(*low+high)/;//要这样找mid
             double mid2=(low+*high)/;//要这样找mid
             sum1=equ(mid1);
             sum2=equ(mid2);
             if(sum1<sum2){
                 low=mid1;
             }
             else{
                 high=mid2;
             }

         }
         if(equ(low)<=y){
             printf("yes\n");
         }
         else{
             printf("no\n");
         }

     }
     return ;
 }