Codeforces Round #339 (Div. 2)-D. Skills
2016-01-15 21:25
344 查看
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 test(s)
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:
题意:给你n个点,算由这n个点围成的多边形绕P点旋转,覆盖的面积是多少?
思路:数学题。先通过坐标变换,转换成绕原点旋转。然后求多边形离原点最远和最近的点,最远的点一定在n个点之中;而最近的点有可能在n个点里面,也有可能在某一条边上。两个圆组成的圆环就是所求的面积
代码
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 test(s)
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:
题意:给你n个点,算由这n个点围成的多边形绕P点旋转,覆盖的面积是多少?
思路:数学题。先通过坐标变换,转换成绕原点旋转。然后求多边形离原点最远和最近的点,最远的点一定在n个点之中;而最近的点有可能在n个点里面,也有可能在某一条边上。两个圆组成的圆环就是所求的面积
代码
[code]#include <iostream> #include <iomanip> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <cmath> #include <algorithm> #define N 100008 #define ll long long #define ull unsigned long long #define pi acos(-1) using namespace std; struct point { ll x, y; void operator -= (point a) { this->x -= a.x; this->y -= a.y; } }node , p; double len(point a, point b) //求线段ab到原点的最小距离 { double r = (a.x-b.x)*(a.x) + (a.y-b.y)*a.y; double d = (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y); if (r <= 0) return a.x*a.x + a.y*a.y; else if (r >= d) return b.x*b.x+b.y*b.y; r /= d; double x = a.x+(b.x-a.x)*r, y = a.y+(b.y-a.y)*r; return x*x+y*y; } int main() { // freopen("1.txt", "r", stdin); ios::sync_with_stdio(false); cin.tie(0); int i, j, n; double r1, r2, t; cin >> n >> p.x >> p.y; r2 = 0; for (i = 0; i < n; i++) { cin >> node[i].x >> node[i].y; node[i] -= p; t = node[i].x*node[i].x + node[i].y*node[i].y; if (r2 < t) r2 = t; } r1 = 9999999999999; node = node[0]; for (i = 0; i < n; i++) r1 = min(r1, len(node[i+1], node[i])); cout.setf(ios::fixed); cout << setprecision(18) << pi*(r2-r1) << endl; return 0; }
相关文章推荐
- 转义符大全
- jquery.form.js使用
- ansible服务模块和组模块使用
- Java语言概述
- codeforces 614B(div.2) 模拟
- 手把手教你自制一寸两寸照
- Sed
- win7 c盘越用越少的原因—win7 c盘空间不够的解决方法
- python list循环语句的使用方法
- 数据库重点知识总结(二)
- IOS动画中的枚举UIViewAnimationOptions
- HDOJ1713(相遇周期)(有点坑)
- Android界面设计的4种方式之二——在Java代码中控制UI界面
- Android截屏(包括未显示部分)
- 总结下自己的入门学黑之路,希望对各位一些帮助!少走点弯路! --写文时间2011-10-26
- grep的妙用
- Codeforces 615 D Multipliers
- UVA 12563:jin ge jin qu hao
- Redis与Memcached的区别
- [Apio2015]巴厘岛的雕塑|动态规划