# Peter and Snow Blower

## 题目描述

Peter got a new snow blower as a New Year present. Of course, Peter decided to try it immediately. After reading the instructions he realized that it does not work like regular snow blowing machines. In order to make it work, you need to tie it to some point that it does not cover, and then switch it on. As a result it will go along a circle around this point and will remove all the snow from its path.
Formally, we assume that Peter’s machine is a polygon on a plane. Then, after the machine is switched on, it will make a circle around the point to which Peter tied it (this point lies strictly outside the polygon). That is, each of the points lying within or on the border of the polygon will move along the circular trajectory, with the center of the circle at the point to which Peter tied his machine.
Peter decided to tie his car to point $P$ and now he is wondering what is the area of the region that will be cleared from snow. Help him.

## 代码

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
const long double pi = 3.1415926535898;
struct point {
long double x, y;
} p;
long double x, y, ma, mi = 1e18;
long long n;
long double get_dist(int s) {
int t = s + 1;
if (t > n) t = 1;
if (p[s].x == p[t].x) return abs(p[s].x);
long double a = (p[t].y - p[s].y) / (p[t].x - p[s].x), b = -1, c = a * p[s].x - p[s].y;
return abs(c) / sqrt(a * a + b * b);
}
long double check(int s) {
int t = s + 1;
if (t > n) t = 1;
long double x1 = p[s].x, y1 = p[s].y, x2 = p[t].x, y2 = p[t].y;
long double p = x1 * (x2 - x1) + y1 * (y2 - y1), q = x2 * (x1 - x2) + y2 * (y1 - y2);
if (p < 0 && q < 0) return get_dist(s) * get_dist(s);
else if (p >= 0) return x1 * x1 + y1 * y1;
else return x2 * x2 + y2 * y2;
}
bool in() {
bool ret = false;
for (long long i = 1, j = n; i <= n; j = i++)
if (((p[i].y > y) != (p[j].y > y)) && ((p[i].x - p[j].x) * p[i].y / (p[j].y - p[i].y) + p[i].x) > 0)
ret = !ret;
return ret;
}
int main()
{
cin >> n >> x >> y;
for (int i = 1; i <= n; ++i) {
cin >> p[i].x >> p[i].y; p[i].x -= x, p[i].y -= y;
ma = max(ma, p[i].x * p[i].x + p[i].y * p[i].y);
if (i > 1) mi = min(mi, check(i - 1));
}
mi = min(mi, check(n)); if (in()) mi = 0;
cout << setprecision(10) << fixed << (ma - mi) * pi << endl;
return 0;
} 418 I'm a teapot