codeforces #339 div 2 C. Peter and Snow Blower

  1/* ***********************************************
  2Author :111qqz
  3Created Time :2016年02月16日 星期二 16时58分59秒
  4File Name :code/cf/problem/614C.cpp
  5************************************************ */
  6
  7#include <cstdio>
  8#include <cstring>
  9#include <iostream>
 10#include <algorithm>
 11#include <vector>
 12#include <queue>
 13#include <set>
 14#include <map>
 15#include <string>
 16#include <cmath>
 17#include <cstdlib>
 18#include <ctime>
 19#define fst first
 20#define sec second
 21#define lson l,m,rt<<1
 22#define rson m+1,r,rt<<1|1
 23#define ms(a,x) memset(a,x,sizeof(a))
 24typedef long long LL;
 25#define pi pair < int ,int >
 26#define MP make_pair
 27
 28using namespace std;
 29const double eps = 1E-8;
 30const int dx4[4]={1,0,0,-1};
 31const int dy4[4]={0,-1,1,0};
 32const int inf = 0x3f3f3f3f;
 33const int N = 1E5+7;
 34const double PI = acos(-1.0);
 35
 36int dblcmp(double d)
 37{
 38    return d<-eps?-1:d>eps;
 39}
 40struct point
 41{
 42    double x,y;
 43    point (){}
 44    point (double _x,double _y):
 45	x(_x),y(_y){};
 46    void input()
 47    {
 48	cin>>x>>y;
 49    }
 50    point sub(point p)
 51    {
 52	return point (x-p.x,y-p.y);
 53    }
 54    double dot(point p)
 55    {
 56	return x*p.x+y*p.y;
 57    }
 58    double distance2(point p)
 59    {
 60	return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y);
 61    }
 62    double distance(point p)
 63    {
 64	return hypot(x-p.x,y-p.y);
 65    }
 66    double det(point p)
 67    {
 68	return x*p.y-y*p.x;
 69    }
 70}p[N],q;
 71double sqr(double x)
 72{
 73    return x*x;
 74}
 75struct line
 76{
 77    point a,b;
 78    line(){}
 79    line (point _a,point _b)
 80    {
 81	a = _a;
 82	b = _b;
 83    }
 84    double length()
 85    {
 86	return a.distance(b);
 87    }
 88
 89    double dispointtoline2(point p)
 90    {
 91	double res = sqr(fabs(p.sub(a).det(b.sub(a)))/length());
 92//	cout<<"res:"<<res<<endl;
 93	return res;
 94    }
 95    double dispointtoseg2(point p)
 96    {
 97	if (dblcmp(p.sub(b).dot(a.sub(b)))<0||dblcmp(p.sub(a).dot(b.sub(a)))<0)
 98	    return min(p.distance2(a),p.distance2(b));
 99
100    return dispointtoline2(p);
101    }
102}li[N];
103
104int n;
105int main()
106{
107	#ifndef  ONLINE_JUDGE 
108	freopen("code/in.txt","r",stdin);
109  #endif
110
111	cin>>n;
112	q.input();
113	double MIN = 99999999999999999;
114	double MAX = -1;
115	for ( int i = 1 ; i <= n ; i++)
116	{
117	    p[i].input();
118	    double d= p[i].distance2(q);
119	    MAX = max(MAX,d);
120	}
121
122	for ( int i = 1 ; i <= n-1 ; i++)
123	{
124	    li[i] = line(p[i],p[i+1]);
125	}
126	li[n]=line(p[n],p[1]);
127	for ( int i = 1 ; i <= n ; i++)
128	{
129	    double d = li[i].dispointtoseg2(q);
130//	    cout<<li[i].a.x<<" "<<li[i].a.y<<"   "<<li[i].b.x<<" "<<li[i].b.y<<endl; 
131	//    cout<<"d:"<<d<<endl;
132	    MIN = min(MIN,d);
133	}
134//	cout<<"MAX:"<<MAX<<"  MIN:"<<MIN<<endl;
135	double ans = (MAX-MIN)*PI;
136	printf("%.18f\n",ans);
137
138
139  #ifndef ONLINE_JUDGE  
140  fclose(stdin);
141  #endif
142    return 0;
143}