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 ...
.NET中 类型,对象,线程栈,托管堆在运行时的关系 The Relationship at Run Time between Types,Objects,A Thread's Stack,and T ...
.Net用户使用期限的设置.限制通用小组件 最近比较项目组的同事都比较烦,不断的穿梭在不同的项目之间,一个人同时要兼顾多个项目的维护修改.甚至刚放下这个客户的电话,另一个客户的电话就进来了.究其原因, ...
C#彻底解决Web Browser 跨域读取Iframes内容 用C# winform的控件web browser 读取网页内容,分析一下数据,做一些采集工作. 如果是同一个域名下面还是好办的,基本上 ...
import java.util.Calendar; import java.util.Date; import java.util.GregorianCalendar; public class M ...
1.Runnable对象 启动线程:(new Thread(new MyRunnable()).start() 2.jsp中<%@ page language="java" ...
public final class String implements java.io.Serializable, Comparable<String>, CharSequence { ...
linux vi编辑常用命令 来源:互联网 作者:佚名 时间:07-10 21:31:14 [大 中 小] linux vi编辑常用命令,需要的朋友可以参考下 vi编辑器中有三种状态模式 1. ...
一.Python的安装 1.下载python安装包https://www.python.org/ 2.选择对应的Python版本(Windows下) 3.装完之后打开电脑的cmd,验证一下安装是否成功 ...
进程相关的概念 程序与进程 程序,是指编译好的二进制文件,在磁盘上,不占用系统资源(CPU.内存.打开的文件.设备.锁等等). 进程,是一个抽象的概念,与操作系统原理联系紧密.进程是活跃的程序,占用系 ...
列出所有端口使用情况 netstat -ano 只查看端口5060使用情况 netstat -ano|findstr "5060" 查看进程8612的信息 tasklist|fin ...