codeforce 399 div1 a

第一次写 多多包含

1.题目:

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:

2.思路

首先,图是由一个图围绕某一个点转出来的,不难想象,图是由无数个点形成对吧,那就等于无数个点围绕着某一个点转一圈,点围绕点所形成的图形必定是一个圆吧,所以说最后的图取决于最外面的点和最里面的点,最外面的点是可以从顶点找到的,这个可以画个图证明,

而最里面的点怎么找?有可能是顶点,也有可能是某条边上的点,这又是为什么呢,还是那句,画个图,作为一名工科类的学生,切忌空想,就举一个简单的例子,一条平行于x轴无限长的直线绕原点转一圈,找最小距离,必定是直线中点到原点的距离,所以说这最短等于高到原点的距离,什么时候高才能用来当作最短呢?就是两个底角都为锐角的时候,这个可以画图证明,如果不是锐角故然会小,但是这个最短是不可取的,因为这条高是作在图的外面,这里可以用秦九公定理.

code:

#include <iostream>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <cmath>  
#include <stack>  
#include <bitset>  
#include <queue>  
#include <set>  
#include <map>  
#include <string>  
#include <algorithm>
#include<queue>
#include<vector>
#include <stdio.h>  
using namespace std;
typedef long long LL;
#define PI acos(-1.0)
struct node {
    double x;
    double y;
}arr[1001000];
int n;
double MAX=0, MIN=100000000;
double dis(node a, node b)
{
    double x = a.x - b.x;
    double y = a.y - b.y;
    return sqrt(x*x + y*y);
}
int main()
{
    cin >> n>>arr[0].x>>arr[0].y;
    for (int i = 1; i <= n; i++)
    {
        cin >> arr[i].x >> arr[i].y;
    }
    for (int i = 1; i <= n; i++)
    {
        MAX = max(MAX, dis(arr[0], arr[i]));
    }
    arr[n + 1] = arr[1];
    for (int i = 1; i <= n; i++)
    {
        double bian1 = dis(arr[i], arr[0]);
        double bian2 = dis(arr[i + 1], arr[0]);
        double bian3 = dis(arr[i], arr[i + 1]);
        if(bian1*bian1>=bian2*bian2+bian3*bian3)
            MIN = min(MIN, bian2);
        else if(bian2*bian2>=bian1*bian1+bian3*bian3)
            MIN = min(MIN, bian1);
        else
        {
            double p = (bian1 + bian2 + bian3) / 2.0;
            double s = sqrt(p*(p - bian1)*(p - bian2)*(p - bian3));
            double h = 2 * s / bian3;
            MIN = min(MIN, h);
        }
    }
    double ans = ((MAX*MAX)-(MIN*MIN))*PI;
    printf("%.7lf", ans);
    return 0;
}