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 ...
E. Axis Walking time limit per test 3 seconds memory limit per test 512 megabytes input standard inp ...
题目地址: http://poj.org/problem?id=2409 给你一串珠子有m个,用n种不同的颜色涂色,问有多少种分法. 用polay定理求解,对于排成一排的带编号的小球,按照某一种方案改 ...
一般来说,提供Web服务的Linux服务器是很少需要连接OpenVPN的,但是个人Linux计算机在很多时候就需要连接OpenVPN了.比如以Linux为开发环境,需要连接公司的OpenVPN等等. ...
1. 首先说一下autocommit 默认情况下autocommit的开关是打开的,也就是ON,查看方法 方法1. select @@[global/session].autocommit; 方法2. ...
创建IDEA项目 单击 Create New Project 按钮,弹出 New Project 对话框. 选择 Maven 选项,单击 Next 按钮. 输入 GroupId.ArtifactId. ...
安装基本环境 1.ubuntu 下载 下载地址:http://releases.ubuntu.com/14.04.4/ 2.安装virtualBox 直接在软件管家中下载即可 3.安装ubuntu 注 ...
1.如果要查询一些库 typings search xxx 2.安装jquery node 这样的库要这样 typings dt~node --global --save 一定要dt~xxx ,然 ...
管道(pipe)本质上是一种文件,管道通信本质上是通过读写文件通信,但是管道解决了文件的两个问题:限制管道大小,解决read()调用文件结束问题. 管道一个环形的缓冲区,通过两个进程以生产者/消费者的 ...
->有这样一个Map集合 Map<String, String> map = new HashMap<String, String>(); map.put(", ...
1.typeof(x)中的x,必须是具体的类名.类型名称等,不可以是变量名称. 2.GetType()方法继承自Object,所以C#中任何对象都具有GetType()方法,它的作用和typeof() ...