HDU 5954 - Do not pour out - [積分+二分][2016ACM/ICPC亞洲區瀋陽站 Problem G]

題目連接：http://acm.hdu.edu.cn/showproblem.php?pid=5954

Problem Description
You have got a cylindrical cup. Its bottom diameter is 2 units and its height is 2 units as well.
The height of liquid level in the cup is d (0 ≤ d ≤ 2). When you incline the cup to the maximal angle such that the liquid inside has not been poured out, what is the area of the surface of the liquid?

Input
The first line is the number of test cases. For each test case, a line contains a float-point number d.

Output
For each test case, output a line containing the area of the surface rounded to 5 decimal places.

Sample Input
4
0
1
2
0.424413182

Sample Output
0.00000
4.44288
3.14159
3.51241

 

題意：

有一個圓柱形杯子，底部直徑爲 $2$，高爲 $2$，告訴你當杯子水平放置時水面高度爲 $d(0 \le d \le 2)$，

求當在水不倒出來的前提下杯子傾斜角度最大時，水面面積。

 

題解：

（參考https://blog.csdn.net/danliwoo/article/details/53002695）

當 $d=1$ 時，是臨界狀況。

當 $d>1$ 時，水面爲一個橢圓，設 $\theta$ 爲水面與杯底的夾角，則 $S = \pi R r = \pi \cdot \frac{1}{cos \theta} \cdot 1 = \frac{\pi}{cos \theta}$。

當 $d<1$ 時，水面爲一個橢圓截取一部分：

若將水此時的形狀，按平行於杯底的方向，分割成若干薄面，每一個薄面的面積爲 $S_0$，則水的體積爲

$V = \int_{0}^{2}S_0dy$；

不難求得

$y_0 = x_0 tan \theta$

$1 + \cos \alpha = x_0$

$S_0 = \pi - \alpha + \sin \alpha \cos \alpha$

上三式，對於 $0 \le \alpha \le \pi$（即 $2 \ge x_0 \ge 0$）均成立。

則水的體積定積分可變爲

$V = \int_{0}^{2}(\pi - \alpha + \sin \alpha \cos \alpha)d[(1 + \cos \alpha)\tan\theta]$

即

$\tan\theta \int_{\pi}^{\alpha_1}(\pi - \alpha + \sin \alpha \cos \alpha)(- \sin \alpha)d\alpha$

其中 $\alpha_1 = \arccos(\frac{2}{\tan\theta}-1)$。

對上式積分得

$V = \tan \theta [(\pi \cos \alpha) + (\sin \alpha - \alpha \cos \alpha) - \frac{1}{3} \sin^3 \alpha]_{\pi}^{\alpha_1}$

那麼，咱們能夠二分 $\theta$，使得 $V$ 逼近 $\pi d$，從而肯定 $\theta$，進而用 $S_{斜面} = \frac{S_{底面}}{cos \theta}$ 求得水面面積。

 

AC代碼：

#include<bits/stdc++.h>
using namespace std;
const double pi=acos(-1.0);
const double eps=1e-12;
inline bool equ(double x,double y){return fabs(x-y)<eps;}
inline double a_t(double t) {
    double tmp=2.0/tan(t);
    if(tmp>2) tmp=2.0;
    else if(tmp<0) tmp=0.0;
    return acos(tmp-1.0);
}
inline double I(double a) {
    return pi*cos(a)+sin(a)-a*cos(a)-pow(sin(a),3)/3.0;
}
inline double V(double t) {
    if(equ(t,pi/2.0)) return 0;
    else return tan(t)*(I(a_t(t))-I(pi));
}
int main()
{
    int T;
    double d;
    cin>>T;
    while(T--)
    {
        cin>>d;
        if(equ(d,0))
        {
            printf("%.5f\n",0);
            continue;
        }
        if(d>1)
        {
            printf("%.5f\n",pi/cos(atan(2-d)));
            continue;
        }

        double l=pi/4.0, r=pi/2.0, t;
        while(r-l>eps)
        {
            t=(l+r)/2;
            if(equ(V(t),pi*d)) break;
            else if(V(t)>pi*d) l=t;
            else r=t;
        }
        double a=a_t(t);
        double S=pi-a+sin(a)*cos(a);
        printf("%.5f\n",S/cos(t));
    }
}