Computational Geometry_Tangling POJ3675 Telescope

Triangulation

Find the area intersection of a circle and a polygon.

Prerequisite knowledge:

Use a vector to find the intersection of a circle and a line.

POJ3675 Telescope

practice:

Divide the polygon into multiple triangles in a counterclockwise direction, connecting the center of the circle to all vertices, and then classify and discuss the intersection of the area of ​​the triangle and the circle.

1. Both A and B are inside the circle

$S_{\cap }=S_{ABC}$

2. Both A and B are outside the circle

$i=arccos(\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|\cdot |\vec{b}|})$

$S_{\cap }=\frac{\theta r^2}{2}$

3.A is inside the circle and B is outside the circle

$S=Sector(C{p}_{b}B)+S_{△AC{p}_b}$

4.A is outside the circle and B is inside the circle

This case is the same as the previous case

5. The special case where both A and B are outside the circle

$S=S_{SectorAC{p}_a}+S_{SectorBC{p}_b}+S_{△p_{a}C{p}_b}$

Therefore, we first find out the two intersection points of the straight line AB and the circle and the shortest distance from the center of the circle to the line segment AB, and then discuss.

1. If both a and b are less than r from the center of the circle, it means the first case

2. If the distance between a and b is greater than r and the distance from the center of the circle to the ab line segment is greater than r, it means the second case

3. Then discuss the length of da and mind in turn to determine whether it is the third or fourth case

4 otherwise the last case.

double circle_triangle(Poi a,Poi b,Cir c={
    
    {
    
    0,0},r}){
    
     //求圆c和三角形的面积交
	auto da=dist(c.p,a),db=dist(c.p,b);
	if(sign(c.r-da)>=0&&sign(c.r-db)>=0) return a*b/2;
	if(!sign(a*b)) return  0.0;
	 //直线ab和圆的交点
	auto mind=poi_to_segment(a,b,c.p);
	// debug(mind);
	vector<Poi> intersection=get_circle_line_intersection(a,b,c);
	Poi pa,pb;
	// if(intersection.size()<2) return 0.0; 
	if(intersection.size()==2)
		pa=intersection[0],pb=intersection[1];
	if(sign(c.r-mind)<=0)	return sector_area(a,b,c);
	if(sign(c.r-da) >= 0) {
    
    //a在圆内，b在圆外
		return (a-c.p)*(pb-c.p)/2.0+sector_area(pb,b,c);
	}
	if(sign(c.r-db) >= 0)
		return sector_area(a,pa,c)+(pa-c.p)*(b-c.p)/2.0;
	return sector_area(a,pa,c)+sector_area(pb,b,c)+(pa-c.p)*(pb-c.p)/2.0;
}

By the way, put my computational geometry template here:

Code:

#include<unordered_set>
#include<unordered_map>
#include<functional>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<iterator>
#include<cstring>
#include<numeric> 
#include<cstdio>
#include<vector>
#include<bitset>
#include<queue>
#include<stack>
#include<cmath>
#include<set>
#include<map>
#define x first
#define y second
using namespace std;
//================================DEBUG
#define FASTIO cin.tie(nullptr) -> sync_with_stdio(false)
#define debug(a) cout << #a": " << a << endl;
#define rep(i, ll, rr) for(int i = ll; i <= rr; ++ i)
#define per(i, rr, ll) for(int i = rr; i >= ll; -- i)
//================================IO
typedef long long LL; typedef unsigned long long ULL; typedef long double LD;
inline LL read(){
    
    LL s=0,w=1;char ch=getchar();for(;!isdigit(ch); ch = getchar())if(ch == '-') w = -1; for (; isdigit(ch);ch = getchar())s=(s<<1)+(s<<3)+(ch^48);return s*w;}
inline void print(LL x,int op=10){
    
    if(!x){
    
    putchar('0');if(op)putchar(op);return;}char F[40];LL tmp=x>0?x:-x;if(x<0)putchar('-');int cnt=0;while(tmp>0){
    
    F[cnt++]=tmp%10+'0';tmp/=10;}while(cnt>0)putchar(F[--cnt]);if(op)putchar(op);}
inline void print128(__int128_t x){
    
    if(x < 0) {
    
    putchar('-');x = -x;}if(x/10) print128(x/10);putchar(x%10+'0');}
template <typename T>void read(T &x){
    
    x=0;int f=1;char ch=getchar();while(!isdigit(ch)){
    
    if(ch=='-')f=-1;ch=getchar();}while(isdigit(ch)){
    
    x=x*10+(ch^48);ch=getchar();}x*=f;return;}
template <typename T>void write(T x){
    
    if(x<0){
    
    putchar('-');x=-x;}if(x>9)write(x/10);putchar(x%10+'0');return;}      
//================================HABIT
LL fpower(LL a,LL b,LL mod) {
    
    LL ans = 1; while(b){
    
     if(b & 1) ans = ans * (a % mod) % mod; a = a % mod * (a % mod) % mod; b >>= 1;} return ans; } 
LL Mod(LL a,LL mod){
    
    return (a%mod+mod)%mod;}
LL gcd(LL a,LL b) {
    
    return b?gcd(b,a%b):a;}
int mov[8][2]={
    
    1,0,0,1,-1,0,0,-1,1,1,-1,-1,1,-1,-1,1};
//================================DEFINE
// #define int LL
#define double long double
typedef pair<int,int> pii;
typedef pair<double,double> pdd;
const double eps=1e-9,PI=acos(-1);
//================================GEOMETRY
int sign(double x) {
    
    if(fabs(x)<eps) return 0;return x>0?1:-1;}
struct Poi{
    
    
	double x,y;
	Poi operator-(Poi b){
    
    return {
    
    x-b.x,y-b.y};}
	Poi operator+(Poi b){
    
    return {
    
    x+b.x,y+b.y};}
	Poi operator*(double k){
    
    return {
    
    x*k,y*k};}
	Poi operator/(double k){
    
    return {
    
    x/k,y/k};}
	Poi norm(){
    
    double d=sqrt(x*x+y*y);return {
    
    x/d,y/d};}
	double operator*(Poi b){
    
    return x*b.y-y*b.x;}
	double operator&(Poi b){
    
    return x*b.x+y*b.y;}
	bool operator==(Poi b){
    
    return sign(x-b.x)==0&&sign(y-b.y)==0;}
	bool operator<(Poi b){
    
    return sign(x-b.x)<0||(sign(x-b.x)==0&&sign(y-b.y)<0);}
};
double cross(Poi a,Poi b){
    
    return a.x*b.y-a.y*b.x;}
double area(Poi a,Poi b,Poi c){
    
    return cross({
    
    b.x-a.x,b.y-a.y},{
    
    c.x-a.x,c.y-a.y});}
double dist(Poi a,Poi b){
    
    double dx=a.x-b.x;double dy=a.y-b.y;return sqrt(dx*dx+dy*dy);}
struct Cir{
    
    Poi p;double r;};
struct Line{
    
    Poi st,ed;};
double get_angle(const Line &a){
    
     //获得直线的角度
	return atan2(a.ed.y-a.st.y,a.ed.x-a.st.x);
}
//直线按照角度的排序函数
bool cmp(Line &a,Line &b){
    
      double A=get_angle(a),B=get_angle(b); if(sign(A-B)==0) return sign(area(a.st,a.ed,b.ed))<0;return A<B;}
//求直线p+kv和直线q+kw的交点
Poi get_line_intersection(Poi p,Poi v,Poi q,Poi w){
    
     auto u=p-q; double t=cross(w,u)/cross(v,w); return {
    
    p.x+v.x*t,p.y+v.y*t};}
//两条线的交点
Poi get_line_intersection(Line a,Line b){
    
     return get_line_intersection(a.st,a.ed-a.st,b.st,b.ed-b.st);}
//bc的交点是否再a的右侧
bool on_right(Line a,Line b,Line c){
    
     auto jiao=get_line_intersection(b,c);return sign(area(a.st,a.ed,jiao))<=0;}
//将一个点顺时针旋转d度
Poi rotate(Poi a,double b){
    
    return {
    
    a.x*cos(b)+a.y*sin(b), -a.x*sin(b)+a.y*cos(b)};}
//获取中垂线
Line get_perpendicular_bisector(Poi a,Poi b){
    
    return {
    
    (a+b)/2,rotate(b-a,PI/2.0)};}
//三点确定圆
Cir get_cir(Poi a,Poi b,Poi c){
    
    auto u=get_perpendicular_bisector(a,b),v=get_perpendicular_bisector(a,c);auto p=get_line_intersection(u.st,u.ed,v.st,v.ed);return {
    
    p, dist(p,a)};}
/*random_suffle(p+1,p+1+n);  点随机化*/
//================================
const int N=200010,M=N*2,mod=1e9+7;
vector<Poi> p;
Poi m,c;int n;
double r;

double len(Poi a){
    
    return sqrt(a&a);}

bool on_segment(Poi p,Poi a,Poi b){
    
     //判断c是否在线段ab上
	return !sign((p-a)*(p-b)) && sign((p-a)&(p-b))<=0;
}

vector<Poi> get_circle_line_intersection(Poi a,Poi b,Cir c={
    
    {
    
    0,0},r}){
    
     //线段ab和圆c的交点
	vector<Poi> ans;
	auto e=get_line_intersection(a, b-a, c.p, rotate(b-a,PI/2)); //弦与中垂线的交点
	auto d=dist(c.p, e); //弦心距
	if(!on_segment(e,a,b)) d=min(dist(c.p,a), dist(c.p, b)); 
	if(sign(c.r-d)<=0) return ans;
	auto len=sqrt(c.r*c.r-dist(c.p, e)*dist(c.p, e));
	Poi pa=e+(a-b).norm()*len,pb=e+((b-a).norm()*len);
	ans.push_back(pa);
	ans.push_back(pb);
	return ans;
}

double poi_to_segment(Poi a,Poi b,Poi c={
    
    0,0}){
    
     //点到线段的距离
	auto e=get_line_intersection(a, b-a, c, rotate(b-a,PI/2)); //弦与中垂线的交点
	auto d=dist(c, e); //弦心距
	if(!on_segment(e,a,b)) d=min(dist(c,a), dist(c, b)); 
	return d;
}

double sector_area(Poi a,Poi b,Cir c){
    
     //c为圆，acb扇形面积
	auto angle=acos((a&b)/len(a)/len(b));
	if(sign(a*b)<0) angle=-angle;
	return c.r*c.r*angle/2.0;
}

double circle_triangle(Poi a,Poi b,Cir c={
    
    {
    
    0,0},r}){
    
     //求圆c和三角形的面积交
	auto da=dist(c.p,a),db=dist(c.p,b);
	if(sign(c.r-da)>=0&&sign(c.r-db)>=0) return a*b/2;
	if(!sign(a*b)) return  0.0;
	 //直线ab和圆的交点
	auto mind=poi_to_segment(a,b,c.p);
	// debug(mind);
	vector<Poi> intersection=get_circle_line_intersection(a,b,c);
	Poi pa,pb;
	// if(intersection.size()<2) return 0.0; 
	if(intersection.size()==2)
		pa=intersection[0],pb=intersection[1];
	if(sign(c.r-mind)<=0)	return sector_area(a,b,c);
	if(sign(c.r-da) >= 0) {
    
    //a在圆内，b在圆外
		return (a-c.p)*(pb-c.p)/2.0+sector_area(pb,b,c);
	}
	if(sign(c.r-db) >= 0)
		return sector_area(a,pa,c)+(pa-c.p)*(b-c.p)/2.0;
	return sector_area(a,pa,c)+sector_area(pb,b,c)+(pa-c.p)*(pb-c.p)/2.0;
}

double area(){
    
    
	double res=0;
	for(int i=0;i<n;++i){
    
    
		res += circle_triangle(p[i], p[(i+1)%n]);
	}
	return fabs(res);
}

void solve(){
    
    
	p.clear();
	p.resize(n);
	for(auto &u:p){
    
    
		scanf("%Lf%Lf",&u.x,&u.y);
	}
	printf("%.2Lf\n",area());
}
//===============================
signed main(){
    
    
	// int _=1;
	while(scanf("%Lf%d",&r,&n)!=-1)	
		solve();
	return 0;
}