HDU 5130 Signal Interference(计算几何 + 模板)
HDU 5130 Signal Interference(计算几何 + 模板)
题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5130
Description
Two countries A-Land and B-Land are at war. The territory of A-Land is a simple polygon with no more than 500 vertices. For military use, A-Land constructed a radio tower (also written as A), and it's so powerful that the whole country was under its signal. To interfere A-Land's communication, B-Land decided to build another radio tower (also written as B). According to an accurate estimation, for any point P, if the euclidean distance between P and B is no more than k (0.2 ≤ k < 0.8) times of the distance between P and A, then point P is not able to receive clear signals from A, i.e. be interfered. Your task is to calculate the area in A-Land's territory that are under B-Land's interference.
Input
There are no more than 100 test cases in the input.
In each test case, firstly you are given a positive integer N indicating the amount of vertices on A-Land's territory, and an above mentioned real number k, which is rounded to 4 digits after the decimal point.
Then N lines follow. Each line contains two integers x and y (|x|, |y| ≤ 1000), indicating a vertex's coordinate on A's territory, in counterclockwise or clockwise order.
The last two lines of a test case give radio tower A and B's coordinates in the same form as vertexes' coordinates. You can assume that A is not equal to B.
Output
For each test case, firstly output the case number, then output your answer in one line following the format shown in sample. Please note that there is a blank after the ':'.
Your solution will be accepted if its absolute error or relative error is no more than 10-6.
This problem is special judged.
Sample Input
4 0.5000
-1 -1
1 -1
1 1
-1 1
0 0
-1 0
Sample Output
Case 1: 0.2729710441
题意：
给你n个点按照顺时针或者逆时针排序围成多边形，A,B点，让你计算从某点到B点的距离是到A距离的K倍，求这个图形和多边形的相交的面积。
题解：
求的点带入，化简就是一个圆，然后就是圆和多边形的面积交。套模板。
代码：
#include <bits/stdc++.h>
#define eps 1e-8
using namespace std;
struct Point{
double x,y;
Point(double x=0, double y=0):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point c,double r):c(c),r(r) {}
Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -1;
if(x > eps) return 1;
return 0;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }
bool OnSegment(Point P, Point A, Point B) {
return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;
}
double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
double DistanceToLine(Point P, Point A, Point B){
Vector v1 = B-A, v2 = P-A;
return fabs(Cross(v1,v2)) / Length(v1);
}
Point DisP(Point A, Point B){
return Length(B-A);
}
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
return max(A.x,B.x) >= min(C.x,D.x) &&
max(C.x,D.x) >= min(A.x,B.x) &&
max(A.y,B.y) >= min(C.y,D.y) &&
max(C.y,D.y) >= min(A.y,B.y) &&
dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 &&
dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0;
}
Point Zero = Point(0,0);
//sum_ans !!!!!!!fabs()
double TriAngleCircleInsection(Circle C, Point A, Point B)
{
Vector OA = A-C.c, OB = B-C.c;
Vector BA = A-B, BC = C.c-B;
Vector AB = B-A, AC = C.c-A;
double DOA = Length(OA), DOB = Length(OB),DAB = Length(AB), r = C.r;
if(dcmp(Cross(OA,OB)) == 0) return 0;
if(dcmp(DOA-C.r) < 0 && dcmp(DOB-C.r) < 0) return Cross(OA,OB)*0.5;
else if(DOB < r && DOA >= r) {
double x = (Dot(BA,BC) + sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-x/DAB)*2/r/DOA)*r*r*0.5+TS*x/DAB;
}
else if(DOB >= r && DOA < r) {
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-y/DAB)*2/r/DOB)*r*r*0.5+TS*y/DAB;
}
else if(fabs(Cross(OA,OB)) >= r*DAB || Dot(AB,AC) <= 0 || Dot(BA,BC) <= 0) {
if(Dot(OA,OB) < 0) {
if(Cross(OA,OB) < 0) return (-acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
else return ( acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
}
else return asin(Cross(OA,OB)/DOA/DOB)*r*r*0.5;
}
else {
double x = (Dot(BA,BC)+sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return (asin(TS*(1-x/DAB)*2/r/DOA)+asin(TS*(1-y/DAB)*2/r/DOB))*r*r*0.5 + TS*((x+y)/DAB-1);
}
}
Point s[600],A,B ;
int main()
{
int n ;
int _t = 0;
while (~scanf("%d",&n)){
double k ;
_t++ ;
scanf("%lf",&k) ;
for (int i = 1;i <= n; i++)
s[i].input();
A.input();B.input();
s[n+1] = s[1];
double D,E,F;
D = (2.0*k*k*A.x - 2.0*B.x)/(1.0-k*k) ;
E = (2.0*k*k*A.y - 2.0*B.y)/(1.0-k*k) ;
F = (B.x*B.x+B.y*B.y-k*k*(A.x*A.x+A.y*A.y))/(1.0-k*k) ;
Circle C = Circle(Point(D*(-0.5),E*(-0.5)),sqrt(D*D+E*E-4.0*F)*0.5) ;
double ans = 0.0;
for (int i = 1; i <= n; i++){
ans = ans + TriAngleCircleInsection(C,s[i],s[i+1]) ;
}
printf("Case %d: %.10lf\n",_t,fabs(ans)) ;
}
return 0;
}
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题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5130
Description
Two countries A-Land and B-Land are at war. The territory of A-Land is a simple polygon with no more than 500 vertices. For military use, A-Land constructed a radio tower (also written as A), and it's so powerful that the whole country was under its signal. To interfere A-Land's communication, B-Land decided to build another radio tower (also written as B). According to an accurate estimation, for any point P, if the euclidean distance between P and B is no more than k (0.2 ≤ k < 0.8) times of the distance between P and A, then point P is not able to receive clear signals from A, i.e. be interfered. Your task is to calculate the area in A-Land's territory that are under B-Land's interference.
Input
There are no more than 100 test cases in the input.
In each test case, firstly you are given a positive integer N indicating the amount of vertices on A-Land's territory, and an above mentioned real number k, which is rounded to 4 digits after the decimal point.
Then N lines follow. Each line contains two integers x and y (|x|, |y| ≤ 1000), indicating a vertex's coordinate on A's territory, in counterclockwise or clockwise order.
The last two lines of a test case give radio tower A and B's coordinates in the same form as vertexes' coordinates. You can assume that A is not equal to B.
Output
For each test case, firstly output the case number, then output your answer in one line following the format shown in sample. Please note that there is a blank after the ':'.
Your solution will be accepted if its absolute error or relative error is no more than 10-6.
This problem is special judged.
Sample Input
4 0.5000
-1 -1
1 -1
1 1
-1 1
0 0
-1 0
Sample Output
Case 1: 0.2729710441
题意：
给你n个点按照顺时针或者逆时针排序围成多边形，A,B点，让你计算从某点到B点的距离是到A距离的K倍，求这个图形和多边形的相交的面积。
题解：
求的点带入，化简就是一个圆，然后就是圆和多边形的面积交。套模板。
代码：
#include <bits/stdc++.h>
#define eps 1e-8
using namespace std;
struct Point{
double x,y;
Point(double x=0, double y=0):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point c,double r):c(c),r(r) {}
Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -1;
if(x > eps) return 1;
return 0;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
Vector VectorUnit(Vector x){ return x / Length(x);}
Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);}
double angle(Vector v) { return atan2(v.y, v.x); }
bool OnSegment(Point P, Point A, Point B) {
return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) < 0;
}
double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
double DistanceToLine(Point P, Point A, Point B){
Vector v1 = B-A, v2 = P-A;
return fabs(Cross(v1,v2)) / Length(v1);
}
Point DisP(Point A, Point B){
return Length(B-A);
}
bool SegmentIntersection(Point A,Point B,Point C,Point D) {
return max(A.x,B.x) >= min(C.x,D.x) &&
max(C.x,D.x) >= min(A.x,B.x) &&
max(A.y,B.y) >= min(C.y,D.y) &&
max(C.y,D.y) >= min(A.y,B.y) &&
dcmp(Cross(C-A,B-A)*Cross(D-A,B-A)) <= 0 &&
dcmp(Cross(A-C,D-C)*Cross(B-C,D-C)) <= 0;
}
Point Zero = Point(0,0);
//sum_ans !!!!!!!fabs()
double TriAngleCircleInsection(Circle C, Point A, Point B)
{
Vector OA = A-C.c, OB = B-C.c;
Vector BA = A-B, BC = C.c-B;
Vector AB = B-A, AC = C.c-A;
double DOA = Length(OA), DOB = Length(OB),DAB = Length(AB), r = C.r;
if(dcmp(Cross(OA,OB)) == 0) return 0;
if(dcmp(DOA-C.r) < 0 && dcmp(DOB-C.r) < 0) return Cross(OA,OB)*0.5;
else if(DOB < r && DOA >= r) {
double x = (Dot(BA,BC) + sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-x/DAB)*2/r/DOA)*r*r*0.5+TS*x/DAB;
}
else if(DOB >= r && DOA < r) {
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return asin(TS*(1-y/DAB)*2/r/DOB)*r*r*0.5+TS*y/DAB;
}
else if(fabs(Cross(OA,OB)) >= r*DAB || Dot(AB,AC) <= 0 || Dot(BA,BC) <= 0) {
if(Dot(OA,OB) < 0) {
if(Cross(OA,OB) < 0) return (-acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
else return ( acos(-1.0)-asin(Cross(OA,OB)/DOA/DOB))*r*r*0.5;
}
else return asin(Cross(OA,OB)/DOA/DOB)*r*r*0.5;
}
else {
double x = (Dot(BA,BC)+sqrt(r*r*DAB*DAB-Cross(BA,BC)*Cross(BA,BC)))/DAB;
double y = (Dot(AB,AC)+sqrt(r*r*DAB*DAB-Cross(AB,AC)*Cross(AB,AC)))/DAB;
double TS = Cross(OA,OB)*0.5;
return (asin(TS*(1-x/DAB)*2/r/DOA)+asin(TS*(1-y/DAB)*2/r/DOB))*r*r*0.5 + TS*((x+y)/DAB-1);
}
}
Point s[600],A,B ;
int main()
{
int n ;
int _t = 0;
while (~scanf("%d",&n)){
double k ;
_t++ ;
scanf("%lf",&k) ;
for (int i = 1;i <= n; i++)
s[i].input();
A.input();B.input();
s[n+1] = s[1];
double D,E,F;
D = (2.0*k*k*A.x - 2.0*B.x)/(1.0-k*k) ;
E = (2.0*k*k*A.y - 2.0*B.y)/(1.0-k*k) ;
F = (B.x*B.x+B.y*B.y-k*k*(A.x*A.x+A.y*A.y))/(1.0-k*k) ;
Circle C = Circle(Point(D*(-0.5),E*(-0.5)),sqrt(D*D+E*E-4.0*F)*0.5) ;
double ans = 0.0;
for (int i = 1; i <= n; i++){
ans = ans + TriAngleCircleInsection(C,s[i],s[i+1]) ;
}
printf("Case %d: %.10lf\n",_t,fabs(ans)) ;
}
return 0;
}
题意: 求所有满足PB <= k*PA 的P所在区域与多边形的交面积. 解法: 2014广州赛区的银牌题,当时竟然没发现是圆,然后就没做出来,然后就gg了. 圆的一般式方程: 设A(x1,y1) ...
//大白p263 #include <cmath> #include <cstdio> #include <cstring> #include <string ...
题意: 给出一个\(n\)个点的简单多边形,和两个点\(A, B\)还有一个常数\(k(0.2 \leq k < 0.8)\). 点\(P\)满足\(\left | PB \right | \l ...
/* HDU5130 Signal Interference http://acm.hdu.edu.cn/showproblem.php?pid=5130 计算几何 圆与多边形面积交 * */ #in ...
整理了一下大白书上的计算几何模板. #include <cstdio> #include <algorithm> #include <cmath> #include ...
题目链接:https://cn.vjudge.net/problem/UVA-12304 题意: 作为题目大合集,有以下一些要求: ①给出三角形三个点,求三角形外接圆,求外接圆的圆心和半径. ②给出三 ...
pro:A的监视区域是一个多边形. 如果A的监视区的内满足到A的距离到不超过到B的距离的K倍的面积大小.K<1 sol:高中几何体经验告诉我们满足题意的区域是个圆,那么就是求圆与多边形的交. # ...
题意:一个很多个点p构成的多边形,pb <= pa * k时p所占区域与多边形相交面积 设p(x,y), (x - xb)^2+(y - yb)^2 / (x - xa)^2+(y ...
Problem Description 小白最近又被空军特招为飞行员,参与一项实战演习.演习的内容还是轰炸某个岛屿(这次的岛屿很大,很大很大很大,大到炸弹怎么扔都能完全在岛屿上引爆),看来小白确实是飞 ...
<script language="javascript" type="text/javascript"> function HideShow() ...
Workflow:自定义工作流 之 模型选择 背景 毕业5年,做了4个版本的工作流框架,工作流几乎是每个企业应用开发人员必须跨过的门槛(我还没有跨过去),下面简要说一下之前的4个版本,然后重点介绍第5 ...
操作系统:WIN10 pro 64 软件版本:VS2015+OpenCV3.0.0 1. 下载安装 http://opencv.org/ https://www.visualstudio.com/ ...
.net分页控件简单实现 好久好久没写博客了.....最近写了一个.net的分页控件,放到园子里...你觉得好,就点个赞,不好呢,就告诉我为啥吧.... 是使用Request.QueryString的 ...
.NET对象占内存多少 一直有一个小小的疑惑——.NET一个对象或者一个集合占多少内存?有没有很快速的方法获取,而不是简单的估计分析对象大小? 查了MSDN,和一些其他人的分析,得到解决是托管代码对象 ...
接着上一篇ajax系列之用jQuery的ajax方法向服务器发出get和post请求写,这篇主要写如何利用ajax和FormData实现页面无刷新的文件上传效果,主要用到了jQuery的ajax()方 ...
在windows上配置JDK 1.下载windows版JDK 网址:http://www.oracle.com/technetwork/java/javase/archive-139210.html ...
预备作业03--我的Linux初体验 学习基于VirtualBox虚拟机安装Ubuntu图文教程在自己笔记本上安装Linux操作系统 一开始以为这个项目很简单,以往也在自己的笔记本上看教程安装过软件, ...
我等蒟蒻爆零之后,问LincHpin大爷:“此等神题可有甚么来头?” LincHpin:“此三题皆为当年ZXR前辈所留.” 固名之,ZXR专场,233~~~ T1 勤奋的YouSiki 这个题在BZO ...
问题简介:算多长时间用本金能得到想要的利息,基础题. 代码:function calculateYears(principal, interest, tax, desired) { // you ...