Codeforces Round #339 (Div. 2)-C（点到线段的最短距离）
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.
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.
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 .
3 0 0 0 1 -1 2 1 2
12.566370614359172464
4 1 -1 0 0 1 2 2 0 1 1
21.991148575128551812
In the first sample snow will be removed from that area:
题意：给你个多边形，再给你个多边形外的点，问你这个多边形绕着这个点旋转所覆盖的面积是多少
题解：我们很容易想到，所覆盖的区域一定是一个圆环形，呢我们求一个最远点和一个最近点就好啦
最远点一定是多边形的一个顶点（很容易想到），但是最近的点却可能是外点到多边形上一条边的垂足。
所以我们再求个点到线段的最短距离即可。
#include<math.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
#define PI 3.1415926
struct node
{
double x,y;
node(){};
node(double a,double b){x=a,y=b;};
node operator - (const node &b) const
{
return node(x-b.x,y-b.y);
}
}a[1000005],s;
//返回一个向量的长度
double Len(node a)
{
return sqrt(a.x*a.x+a.y*a.y);
}
double d(node a,node b)
{
return Len(node(a-b));
}
//求两个向量的点积
double Dot(node a,node b)
{
return a.x*b.x+a.y*b.y;
}
//返回点a到线段bc的最短距离
double PointToSegment(node a,node b,node c)
{
node x=a-b;
node y=c-b;
node z=c-a;
if(Dot(x,y)<0)return Len(x);
if(Dot(y,z)<0)return Len(z);
return fabs((x.x*y.y-x.y*y.x)/Len(y));
}
int main(void)
{
int n,i,j;
scanf("%d%lf%lf",&n,&s.x,&s.y);
for(i=1;i<=n;i++)
scanf("%lf%lf",&a[i].x,&a[i].y);
a[n+1]=a[1];
double maxs=0,mins=1e18;
for(i=1;i<=n;i++)
{
maxs=max(maxs,d(a[i],s));
mins=min(mins,PointToSegment(s,a[i],a[i+1]));
}
double ans=PI*(maxs*maxs-mins*mins);
printf("%.15f\n",ans);
return 0;
}