Project Euler 363 Bézier Curves（几何+二分）

题目链接：

https://projecteuler.net/problem=363

题目：

A cubic Bézier curve is defined by four points: \(P_0, P_1, P_2\) and \(P_3\).

The curve is constructed as follows:

On the segments \(P_0P_1\), \(P_1P_2\) and \(P_2P_3\) the points \(Q_0,Q_1\) and \(Q_2\) are drawn such that

\(P_0Q_0 / P_0P_1 = P_1Q_1 / P_1P_2 = P_2Q_2 / P_2P_3 = t (t in [0,1]).\)

On the segments \(Q_0Q_1\) and \(Q_1Q_2\) the points \(R_0\) and \(R_1\) are drawn such that

\(Q_0R_0 / Q_0Q_1 = Q_1R_1 / Q_1Q_2 = t\) for the same value of \(t\).

On the segment \(R_0R_1\) the point B is drawn such that \(R_0B / R_0R1 = t\) for the same value of \(t\).

The Bézier curve defined by the points \(P_0, P_1, P_2, P_3\) is the locus of B as \(Q_0\) takes all possible positions on the segment \(P_0P_1\).

(Please note that for all points the value of t is the same.)

At this (external) web address you will find an applet that allows you to drag the points \(P_0, P_1, P_2 and P_3\) to see what the Bézier curve (green curve) defined by those points looks like. You can also drag the point \(Q_0\) along the segment \(P_0P_1\).

From the construction it is clear that the Bézier curve will be tangent to the segments \(P_0P_1\) in \(P_0\) and \(P_2P_3\) in \(P_3\).

A cubic Bézier curve with $P_0=(1,0), P_1=(1,v), P_2=(v,1) and P_3=(0,1) $is used to approximate a quarter circle.

The value v > 0 is chosen such that the area enclosed by the lines \(OP_0, OP_3\) and the curve is equal to π/4 (the area of the quarter circle).

By how many percent does the length of the curve differ from the length of the quarter circle?

That is, if L is the length of the curve, calculate 100 × (L − π/2)/(π/2)

Give your answer rounded to 10 digits behind the decimal point.

题解：

这道题超级有意思。

不查一下wiki根本不知道怎么下手...

https://en.wikipedia.org/wiki/Bézier_curve

因为Bézier curve上的4个点是：\((1, 0), (1, v), (v, 1), (0, 1)\)

所以从wiki上得Specific cases可以知道，将4个点代入 \(B(t)\) 得到 \(x(t)\)和 \(y(t)\)：

+ - \(B(t) = (1-t)^3 P_0 + 3(1-t)^2 t P_1 + 3(1-t) t^2 P_2 + t^3 P_3\)

+ - \(x(t) = (1 - t)^3 + 3(1-t)^2t + 3(1-t) t^2 v\)

+ - \(y(t) = 3(1-t)^2tv + 3(1-t) t^2 + t^3\)

其中，\(v\) 未知，\(0 <= t <= 1\). \(x(t), y(t)\) 就是计算曲线上的坐标。

然后我们直接二分 \(v\) 就可以了。

得到 \(v\) 后可以直接计算曲线的长度(arc_length_of_a_curve)，也可以将曲线化为无数的小线段进行逼近。

代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int maxn = 1e8;
const int mod = 1e9;
const double pi = acos(-1.0);

double x(double t, double v)
{
    return (1-t)*(1-t)*(1-t) + 3*(1-t)*(1-t)*t + 3*(1-t)*t*t*v;
}

double y(double t, double v)
{
    return 3*t*(1-t)*(1-t)*v + t*t*t + 3*t*t*(1-t);
}

double cal_distance(double a,double b,double c,double d)
{
  return (double)sqrt((a-c)*(a-c)+(b-d)*(b-d));
}

//https://en.wikipedia.org/wiki/B%C3%A9zier_curve
int main(int argc, char const *argv[]) {

  int dot =  500000;
  double left = 0.0, right = 1.0;
  double v = 0.0, area = 0.0;
  while (right - left > 1e-15) {
     v = (right + left) / 2;
     area = 0.0;
    for(int i = 0;i < dot;i++) {
      double now = (double) i / dot * 1.0;
      double next = (double) (i+1) / dot * 1.0;
      //cross product
      area += ( x(now,v) * y(next,v) - x(next,v) * y(now,v) ) / 2.0;
    }
    if(area > pi / 4.0) {
      right = v;
    }
    else {
      left = v;
    }
  }

  std::cout << "binary search finish !" << '\n';
  std::cout << "left = " << left << '\n';
  std::cout << "v = " << v << '\n';
  std::cout << "area = " << area << '\n';

  double L = 0.0;
  dot = 500000;
  for(int i = 0;i < dot;i++){
    double now = (double)i / dot * 1.0;
    double next = (double)(i + 1) / dot * 1.0;
    L += cal_distance(x(now, v), y(now, v), x(next, v), y(next, v));
  }
  printf("%.12f\n",100.0*(L-pi/2.0)/pi*2.0);
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}