HDU2438 Turn the corner【三分法】【数学几何】

Turn the corner

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

Total Submission(s): 1930    Accepted Submission(s): 736

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

题目大意：有一个直角拐角，给你水平道路宽度Y和竖直高度X，再给你汽车的长l，宽w

问：汽车是否能通过这个拐角。

思路：如果汽车的宽度大于水平道路宽度Y或是竖直高度X，无论如何都通不过。接下来

考虑一般情况。

如图：若汽车最左边与墙一直靠紧，则只需要判断右边最高点是否超过了Y。

设θ为汽车与水平方向的夹角，s为汽车最右边的角到拐点的水平距离。那么

s = l*cos(θ) + w*sin(θ) - x，从而得出 h = s*tan(θ)+w*cos(θ)。

θ角从0~π/2,变化，h则从低到高再到底，且是一个凸形函数。利用三分方法

得到最高点的h，与Y比较判断是否能通过。

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>using namespace std;const double PI = acos(-1.0);double x,y,l,w;double calc(double angle){    double s = l*cos(angle) + w*sin(angle) - x;    double h = s*tan(angle) + w*cos(angle);    return h;}int main(){    while(cin >> x >> y >> l >> w)    {        double left,right,mid,midmid;        left = 0;        right = PI/2;        while(right-left >= 1e-7)        {            mid = (left+right)/2;            midmid = (mid+right)/2;            if(calc(mid) > calc(midmid))                right = midmid;            else                left = mid;        }        if(x<w || y<w || calc(mid) > y)            cout << "no" << endl;        else            cout << "yes" << endl;    }    return 0;}