Codeforces Round #339 (Div. 1) A. Peter and Snow Blower 计算几何

A. Peter and Snow Blower

题目连接：

http://www.codeforces.com/contest/613/problem/A

Description

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 .

Sample Input

3 0 0

0 1

-1 2

1 2

Sample Output

12.566370614359172464

Hint

题意

给你一个多边形，然后给你一个原点，问你这个多边形绕着这个原点旋转一圈所经过的面积是多少

原点一定不在多边形内

题解：

很显然是一个大圆减去一个小圆的面积，所以我们直接找最远的距离和最近的距离就好了

最远的距离一定是原点到某个点的距离，但是最近的就不一定了

注意，多边形，不一定是一个凸包哦

代码

#include<bits/stdc++.h>
using namespace std;
struct point
{
    double x,y;
};
const double PI = acos(-1.0);
point po[100005];
int n;
point T;
double dis(point a,point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double Dot(point a,point b) //点积
{
    return a.x*b.x+a.y*b.y;
}
double Length(point a){return sqrt(Dot(a,a));} //向量的长度
double point_to_seg(point a,point b,point c)  //点a到线段bc的距离
{
    point v1,v2,v3;
    v1.x=c.x-b.x;
    v1.y=c.y-b.y;
    v2.x=a.x-b.x;
    v2.y=a.y-b.y;
    v3.x=a.x-c.x;
    v3.y=a.y-c.y;
    if(Dot(v1,v2)<0) return Length(v2);
    else if(Dot(v1,v3)>0) return Length(v3);
    else return fabs((v1.x*v2.y-v2.x*v1.y)/Length(v1));
}
int main()
{
    scanf("%d",&n);
    scanf("%lf%lf",&T.x,&T.y);
    for(int i=0;i<n;i++)
        scanf("%lf%lf",&po[i].x,&po[i].y);
    po[n].x=po[0].x,po[n].y=po[0].y;
    double Max1 = 0;
    double Min1 = 1000000000.0;
    for(int i=0;i<n;i++)
    {
        Max1 = max(Max1,dis(T,po[i]));
        Min1 = min(Min1,point_to_seg(T,po[i],po[i+1]));
    }
    printf("%.10f\n",Max1*Max1*PI-Min1*Min1*PI);
}