http://codeforces.com/contest/614/problem/C#
题意：给一个多边形和多边形外一定点，多边形绕定点旋转，问多边形扫过的面积。 思路：简单计算几何，找到多边形距离定点的最大和最小距离R和r，答案就是(R^2-R^2)*PI 需要注意的是：最大距离一定是从某点上取得，但是最小距离可能不在顶点上，而在某条边上。
代码实现
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}