codeforces 613A Peter and Snow Blower 点到线段距离

http://codeforces.com/problemset/problem/613/A
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.

Input

The first line of the input contains three integers — the number of vertices of the polygon n (), and coordinates of point P.

Each of the next n lines contains two integers — coordinates of the vertices of the polygon in the clockwise or counterclockwise order. It is guaranteed that no three consecutive vertices lie on a common straight line.

All the numbers in the input are integers that do not exceed 1 000 000 in their absolute value.

Output

Print a single real value number — the area of the region that will be cleared. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Examples
Input

3 0 0
0 1
-1 2
1 2

Output

12.566370614359172464

Input

4 1 -1
0 0
1 2
2 0
1 1

Output

21.991148575128551812

Note

In the first sample snow will be removed from that area:

题目大意：给出一个多边形以及一个点的坐标，求以这个为圆心覆盖给定多边形的圆环的最小面积。
思路：很明显，所求的圆环面积可以用一个大圆的面积减去一个小圆的面积。大圆半径$R$=圆心到多边形的最远距离，小圆半径$r$=圆心到多边形的最小距离。仔细思考不难发现，最远距离一定是圆心到多边形某个顶点的距离，而最近距离就不一定是点到点了，可能是圆心到多边形某条边的距离。那么我们枚举多边形的$n$条边计算圆心到它们的距离就行了。

#include
#include
#include
#include
#include
#include
using namespace std;

const double pi=acos(-1.0);//弧度pi
const double eps=1e-10;//精度

struct point
{
    double x,y;
    point(double a=0,double b=0)
    {
        x=a,y=b;
    }
    friend point operator * (point a,double b)
    {
        return point(a.x*b,a.y*b);
    }
    friend point operator * (double a,point b)
    {
        return point(b.x*a,b.y*a);
    }
    point operator - (const point &b)const
    {
        return point(x-b.x,y-b.y);
    }
    point operator + (const point &b)const
    {
        return point(x+b.x,y+b.y);
    }
    point operator / (const double b)const
    {
        return point(x/b,y/b);
    }
    bool operator < (const point &b)const//按坐标排序
    {
        if(fabs(x-b.x)<eps)
            return y<b.y-eps;
        return x<b.x-eps;
    }
    void transxy(double sinb,double cosb)//逆时针旋转b弧度
    {                                      //若顺时针 在传入的sinb前加个-即可
        double tx=x,ty=y;
        x=tx*cosb-ty*sinb;
        y=tx*sinb+ty*cosb;
    }
    void transxy(double b)//逆时针旋转b弧度
    {                     //若顺时针传入-b即可
        double tx=x,ty=y;
        x=tx*cos(b)-ty*sin(b);
        y=tx*sin(b)+ty*cos(b);
    }
    double norm()
    {
        return sqrt(x*x+y*y);
    }
};

inline double dot(point a,point b)//点积
{
    return a.x*b.x+a.y*b.y;
}
inline double cross(point a,point b)//叉积
{
    return a.x*b.y-a.y*b.x;
}

inline double dist(point a,point b)//两点间距离
{
    return (a-b).norm();
}

inline int sgn(double x)
{
    if(fabs(x)<eps)
        return 0;
    if(x>0)
        return 1;
    return -1;
}

typedef point Vector;

struct line
{
    point s,e;
    line(){}
    line(point _s,point _e)
    {
        s=_s,e=_e;
    }
    bool operator <(const line &a)const
    {
        return 1;
    }
};

inline double dis_point_seg(point p,line l)
{
    if(sgn(dot(p-l.s,l.e-l.s))<0)//不存在 返回点到线段端点的距离
        return (p-l.s).norm();
    if(sgn(dot(p-l.e,l.s-l.e))<0)//不存在 返回点到线段端点的距离
        return (p-l.e).norm();
    return fabs(cross(l.s-p,l.e-p))/dist(l.s,l.e);//|叉积|=2*S△
}

int n;
point pt[100005];

int main()
{
    point tmp;
    double MAX=-1e16,MIN=1e16;
    scanf("%d%lf%lf",&n,&tmp.x,&tmp.y);
    for(int i=1;i<=n;i++)
    {
        scanf("%lf%lf",&pt[i].x,&pt[i].y);
        MAX=max(MAX,(pt[i]-tmp).norm());
    }
    for(int i=2;i<=n;i++)
        MIN=min(MIN,dis_point_seg(tmp,line(pt[i],pt[i-1])));
    MIN=min(MIN,dis_point_seg(tmp,line(pt[1],pt[n])));
    printf("%.16f\n",pi*(MAX*MAX-MIN*MIN));
    return 0;
}