Codeforces614C Peter and Snow Blower

题意：各一个点，一个多边形，多边形绕着这个点转，问转过的面积是多少

题解：可以发现转过的面积就是一个圆环，这里的多边形可能是凹多边形，环外的半径就是到点的最远距离，环内的半径就是多边形的边的线段到点的最小距离

#include <bits/stdc++.h>
#define maxn 100010
#define INF 0x3f3f3f3f
typedef long long ll;
using namespace std;
double eps = 1e-6;
#define Vector Point
int dcmp(double x) { return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1); }
struct Point {
    double x, y;

    Point(const Point& rhs): x(rhs.x), y(rhs.y) { } //拷贝构造函数
    Point(double x = 0.0, double y = 0.0): x(x), y(y) { }   //构造函数

    friend istream& operator >> (istream& in, Point& P) { return in >> P.x >> P.y; }
    friend ostream& operator << (ostream& out, const Point& P) { return out << P.x << ' ' << P.y; }

    friend Vector operator + (const Vector& A, const Vector& B) { return Vector(A.x+B.x, A.y+B.y); }
    friend Vector operator - (const Point& A, const Point& B) { return Vector(A.x-B.x, A.y-B.y); }
    friend Vector operator * (const Vector& A, const double& p) { return Vector(A.x*p, A.y*p); }
    friend Vector operator / (const Vector& A, const double& p) { return Vector(A.x/p, A.y/p); }
    friend bool operator == (const Point& A, const Point& B) { return dcmp(A.x-B.x) == 0 && dcmp(A.y-B.y) == 0; }
    friend bool operator < (const Point& A, const Point& B) { return A.x < B.x || (A.x == B.x && A.y < B.y); }

    void in(void) { scanf("%lf%lf", &x, &y); }
    void out(void) { printf("%lf %lf", x, y); }
}a[maxn];
double Dot(const Vector& A, const Vector& B) { return A.x*B.x + A.y*B.y; }  //点积
double Length(const Vector& A){ return sqrt(Dot(A, A)); }
double Angle(const Vector& A, const Vector& B) { return acos(Dot(A, B)/Length(A)/Length(B)); }  //向量夹角
double Cross(const Vector& A, const Vector& B) { return A.x*B.y - A.y*B.x; }    //叉积
double Area(const Point& A, const Point& B, const Point& C) { return fabs(Cross(B-A, C-A)); }
double f(const Point& P, const Point& A, const Point& B) {
    if (A == B) return Length(P - A);
    Vector v1 = B - A, v2 = P - A, v3 = P - B;
    if (dcmp(Dot(v1, v2)) < 0) return Length(v2);
    if (dcmp(Dot(v1, v3)) > 0) return Length(v3);
    return fabs(Cross(v1, v2)) / Length(v1);
}
int main(){
    double x, y, x1, y1, ma = 0, mi=1e18;
    int n;
    scanf("%d%lf%lf", &n, &a[0].x, &a[0].y);
    for(int i=2;i<=n+1;i++)
        scanf("%lf%lf", &a[i].x, &a[i].y);
    a[1] = a[n+1];
    for(int i=2;i<=n+1;i++){
        double t = f(a[0], a[i], a[i-1]);
        mi = min(t, mi);
        ma = max(ma, Length(a[i]-a[0]));
    }
    printf("%.9f\n", (ma*ma-mi*mi)*acos(-1.0));
    return 0;
}