POJ 3675 Telescope(简单多边形和圆的面积交)
Description
Updog is watching a plane object with a telescope. The field of vision in the telescope can be described as a circle. The center is at the origin and the radius is R. The object can be seen as a simple polygon of N vertexes. Updog wants to know the area of the part of the object that can be seen in the telescope.
Input
The input will contain several test cases. For each case:
The first line contains only one real number R.
The second line contains an integer N. The following N lines contain two real numbers xi and yi each, which describe the coordinates of a vertex. Two vertexes in adjacent lines are also adjacent on the polygon.
Constraints: 3 ≤ N ≤50, 0.1 ≤ R ≤1000, -1000 ≤ xi, yi ≤ 1000
Output
For each test case, output one real number on a separate line, which is the area of the part that can be seen. The result should be rounded to two decimal places.
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
#define sqr(x) ((x) * (x)) const int MAXN = ;
const double EPS = 1e-;
const double PI = acos(-1.0);//3.14159265358979323846
const double INF = ; inline int sgn(double x) {
return (x > EPS) - (x < -EPS);
} struct Point {
double x, y, ag;
Point() {}
Point(double x, double y): x(x), y(y) {}
void read() {
scanf("%lf%lf", &x, &y);
}
bool operator == (const Point &rhs) const {
return sgn(x - rhs.x) == && sgn(y - rhs.y) == ;
}
bool operator < (const Point &rhs) const {
if(y != rhs.y) return y < rhs.y;
return x < rhs.x;
}
Point operator + (const Point &rhs) const {
return Point(x + rhs.x, y + rhs.y);
}
Point operator - (const Point &rhs) const {
return Point(x - rhs.x, y - rhs.y);
}
Point operator * (const double &b) const {
return Point(x * b, y * b);
}
Point operator / (const double &b) const {
return Point(x / b, y / b);
}
double operator * (const Point &rhs) const {
return x * rhs.x + y * rhs.y;
}
double length() {
return sqrt(x * x + y * y);
}
double angle() {
return atan2(y, x);
}
Point unit() {
return *this / length();
}
void makeAg() {
ag = atan2(y, x);
}
void print() {
printf("%.10f %.10f\n", x, y);
}
};
typedef Point Vector; double dist(const Point &a, const Point &b) {
return (a - b).length();
} double cross(const Point &a, const Point &b) {
return a.x * b.y - a.y * b.x;
}
//ret >= 0 means turn right
double cross(const Point &sp, const Point &ed, const Point &op) {
return cross(sp - op, ed - op);
} double area(const Point& a, const Point &b, const Point &c) {
return fabs(cross(a - c, b - c)) / ;
}
//counter-clockwise
Point rotate(const Point &p, double angle, const Point &o = Point(, )) {
Point t = p - o;
double x = t.x * cos(angle) - t.y * sin(angle);
double y = t.y * cos(angle) + t.x * sin(angle);
return Point(x, y) + o;
} double cosIncludeAngle(const Point &a, const Point &b, const Point &o) {
Point p1 = a - o, p2 = b - o;
return (p1 * p2) / (p1.length() * p2.length());
} double includedAngle(const Point &a, const Point &b, const Point &o) {
return acos(cosIncludeAngle(a, b, o));
/*
double ret = abs((a - o).angle() - (b - o).angle());
if(sgn(ret - PI) > 0) ret = 2 * PI - ret;
return ret;
*/
} struct Seg {
Point st, ed;
double ag;
Seg() {}
Seg(Point st, Point ed): st(st), ed(ed) {}
void read() {
st.read(); ed.read();
}
void makeAg() {
ag = atan2(ed.y - st.y, ed.x - st.x);
}
};
typedef Seg Line; //ax + by + c > 0
Line buildLine(double a, double b, double c) {
if(sgn(a) == && sgn(b) == ) return Line(Point(sgn(c) > ? - : , INF), Point(, INF));
if(sgn(a) == ) return Line(Point(sgn(b), -c/b), Point(, -c/b));
if(sgn(b) == ) return Line(Point(-c/a, ), Point(-c/a, sgn(a)));
if(b < ) return Line(Point(, -c/b), Point(, -(a + c) / b));
else return Line(Point(, -(a + c) / b), Point(, -c/b));
} void moveRight(Line &v, double r) {
double dx = v.ed.x - v.st.x, dy = v.ed.y - v.st.y;
dx = dx / dist(v.st, v.ed) * r;
dy = dy / dist(v.st, v.ed) * r;
v.st.x += dy; v.ed.x += dy;
v.st.y -= dx; v.ed.y -= dx;
} bool isOnSeg(const Seg &s, const Point &p) {
return (p == s.st || p == s.ed) ||
(((p.x - s.st.x) * (p.x - s.ed.x) < ||
(p.y - s.st.y) * (p.y - s.ed.y) < ) &&
sgn(cross(s.ed, p, s.st)) == );
} bool isInSegRec(const Seg &s, const Point &p) {
return sgn(min(s.st.x, s.ed.x) - p.x) <= && sgn(p.x - max(s.st.x, s.ed.x)) <=
&& sgn(min(s.st.y, s.ed.y) - p.y) <= && sgn(p.y - max(s.st.y, s.ed.y)) <= ;
} bool isIntersected(const Point &s1, const Point &e1, const Point &s2, const Point &e2) {
return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
(max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
(max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
(max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
(cross(s2, e1, s1) * cross(e1, e2, s1) >= ) &&
(cross(s1, e2, s2) * cross(e2, e1, s2) >= );
} bool isIntersected(const Seg &a, const Seg &b) {
return isIntersected(a.st, a.ed, b.st, b.ed);
} bool isParallel(const Seg &a, const Seg &b) {
return sgn(cross(a.ed - a.st, b.ed - b.st)) == ;
} //return Ax + By + C =0 's A, B, C
void Coefficient(const Line &L, double &A, double &B, double &C) {
A = L.ed.y - L.st.y;
B = L.st.x - L.ed.x;
C = L.ed.x * L.st.y - L.st.x * L.ed.y;
}
//point of intersection
Point operator * (const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
Point I;
I.x = - (B2 * C1 - B1 * C2) / (A1 * B2 - A2 * B1);
I.y = (A2 * C1 - A1 * C2) / (A1 * B2 - A2 * B1);
return I;
} bool isEqual(const Line &a, const Line &b) {
double A1, B1, C1;
double A2, B2, C2;
Coefficient(a, A1, B1, C1);
Coefficient(b, A2, B2, C2);
return sgn(A1 * B2 - A2 * B1) == && sgn(A1 * C2 - A2 * C1) == && sgn(B1 * C2 - B2 * C1) == ;
} double Point_to_Line(const Point &p, const Line &L) {
return fabs(cross(p, L.st, L.ed)/dist(L.st, L.ed));
} double Point_to_Seg(const Point &p, const Seg &L) {
if(sgn((L.ed - L.st) * (p - L.st)) < ) return dist(p, L.st);
if(sgn((L.st - L.ed) * (p - L.ed)) < ) return dist(p, L.ed);
return Point_to_Line(p, L);
} double Seg_to_Seg(const Seg &a, const Seg &b) {
double ans1 = min(Point_to_Seg(a.st, b), Point_to_Seg(a.ed, b));
double ans2 = min(Point_to_Seg(b.st, a), Point_to_Seg(b.ed, a));
return min(ans1, ans2);
} struct Circle {
Point c;
double r;
Circle() {}
Circle(Point c, double r): c(c), r(r) {}
void read() {
c.read();
scanf("%lf", &r);
}
double area() const {
return PI * r * r;
}
bool contain(const Circle &rhs) const {
return sgn(dist(c, rhs.c) + rhs.r - r) <= ;
}
bool contain(const Point &p) const {
return sgn(dist(c, p) - r) <= ;
}
bool intersect(const Circle &rhs) const {
return sgn(dist(c, rhs.c) - r - rhs.r) < ;
}
bool tangency(const Circle &rhs) const {
return sgn(dist(c, rhs.c) - r - rhs.r) == ;
}
Point pos(double angle) const {
Point p = Point(c.x + r, c.y);
return rotate(p, angle, c);
}
}; double CommonArea(const Circle &A, const Circle &B) {
double area = 0.0;
const Circle & M = (A.r > B.r) ? A : B;
const Circle & N = (A.r > B.r) ? B : A;
double D = dist(M.c, N.c);
if((D < M.r + N.r) && (D > M.r - N.r)) {
double cosM = (M.r * M.r + D * D - N.r * N.r) / (2.0 * M.r * D);
double cosN = (N.r * N.r + D * D - M.r * M.r) / (2.0 * N.r * D);
double alpha = * acos(cosM);
double beta = * acos(cosN);
double TM = 0.5 * M.r * M.r * (alpha - sin(alpha));
double TN = 0.5 * N.r * N.r * (beta - sin(beta));
area = TM + TN;
}
else if(D <= M.r - N.r) {
area = N.area();
}
return area;
} int intersection(const Seg &s, const Circle &cir, Point &p1, Point &p2) {
double angle = cosIncludeAngle(s.ed, cir.c, s.st);
//double angle1 = cos(includedAngle(s.ed, cir.c, s.st));
double B = dist(cir.c, s.st);
double a = , b = - * B * angle, c = sqr(B) - sqr(cir.r);
double delta = sqr(b) - * a * c;
if(sgn(delta) < ) return ;
if(sgn(delta) == ) delta = ;
double x1 = (-b - sqrt(delta)) / ( * a), x2 = (-b + sqrt(delta)) / ( * a);
Vector v = (s.ed - s.st).unit();
p1 = s.st + v * x1;
p2 = s.st + v * x2;
return + sgn(delta);
} double CommonArea(const Circle &cir, Point p1, Point p2) {
if(p1 == cir.c || p2 == cir.c) return ;
if(cir.contain(p1) && cir.contain(p2)) {
return area(cir.c, p1, p2);
} else if(!cir.contain(p1) && !cir.contain(p2)) {
Point q1, q2;
int t = intersection(Line(p1, p2), cir, q1, q2);
if(t == ) {
double angle = includedAngle(p1, p2, cir.c);
return 0.5 * sqr(cir.r) * angle;
} else {
double angle1 = includedAngle(p1, p2, cir.c);
double angle2 = includedAngle(q1, q2, cir.c);
if(isInSegRec(Seg(p1, p2), q1))return 0.5 * sqr(cir.r) * (angle1 - angle2 + sin(angle2));
else return 0.5 * sqr(cir.r) * angle1;
}
} else {
if(cir.contain(p2)) swap(p1, p2);
Point q1, q2;
intersection(Line(p1, p2), cir, q1, q2);
double angle = includedAngle(q2, p2, cir.c);
double a = area(cir.c, p1, q2);
double b = 0.5 * sqr(cir.r) * angle;
return a + b;
}
} struct Triangle {
Point p[];
Triangle() {}
Triangle(Point *t) {
for(int i = ; i < ; ++i) p[i] = t[i];
}
void read() {
for(int i = ; i < ; ++i) p[i].read();
}
double area() const {
return ::area(p[], p[], p[]);
}
Point& operator[] (int i) {
return p[i];
}
}; double CommonArea(Triangle tir, const Circle &cir) {
double ret = ;
ret += sgn(cross(tir[], cir.c, tir[])) * CommonArea(cir, tir[], tir[]);
ret += sgn(cross(tir[], cir.c, tir[])) * CommonArea(cir, tir[], tir[]);
ret += sgn(cross(tir[], cir.c, tir[])) * CommonArea(cir, tir[], tir[]);
return abs(ret);
} struct Poly {
int n;
Point p[MAXN];//p[n] = p[0]
void init(Point *pp, int nn) {
n = nn;
for(int i = ; i < n; ++i) p[i] = pp[i];
p[n] = p[];
}
double area() {
if(n < ) return ;
double s = p[].y * (p[n - ].x - p[].x);
for(int i = ; i < n; ++i)
s += p[i].y * (p[i - ].x - p[i + ].x);
return s / ;
}
};
//the convex hull is clockwise
void Graham_scan(Point *p, int n, int *stk, int &top) {//stk[0] = stk[top]
sort(p, p + n);
top = ;
stk[] = ; stk[] = ;
for(int i = ; i < n; ++i) {
while(top && cross(p[i], p[stk[top]], p[stk[top - ]]) <= ) --top;
stk[++top] = i;
}
int len = top;
stk[++top] = n - ;
for(int i = n - ; i >= ; --i) {
while(top != len && cross(p[i], p[stk[top]], p[stk[top - ]]) <= ) --top;
stk[++top] = i;
}
}
//use for half_planes_cross
bool cmpAg(const Line &a, const Line &b) {
if(sgn(a.ag - b.ag) == )
return sgn(cross(b.ed, a.st, b.st)) < ;
return a.ag < b.ag;
}
//clockwise, plane is on the right
bool half_planes_cross(Line *v, int vn, Poly &res, Line *deq) {
int i, n;
sort(v, v + vn, cmpAg);
for(i = n = ; i < vn; ++i) {
if(sgn(v[i].ag - v[i-].ag) == ) continue;
v[n++] = v[i];
}
int head = , tail = ;
deq[] = v[], deq[] = v[];
for(i = ; i < n; ++i) {
if(isParallel(deq[tail - ], deq[tail]) || isParallel(deq[head], deq[head + ]))
return false;
while(head < tail && sgn(cross(v[i].ed, deq[tail - ] * deq[tail], v[i].st)) > )
--tail;
while(head < tail && sgn(cross(v[i].ed, deq[head] * deq[head + ], v[i].st)) > )
++head;
deq[++tail] = v[i];
}
while(head < tail && sgn(cross(deq[head].ed, deq[tail - ] * deq[tail], deq[head].st)) > )
--tail;
while(head < tail && sgn(cross(deq[tail].ed, deq[head] * deq[head + ], deq[tail].st)) > )
++head;
if(tail <= head + ) return false;
res.n = ;
for(i = head; i < tail; ++i)
res.p[res.n++] = deq[i] * deq[i + ];
res.p[res.n++] = deq[head] * deq[tail];
res.n = unique(res.p, res.p + res.n) - res.p;
res.p[res.n] = res.p[];
return true;
} //ix and jx is the points whose distance is return, res.p[n - 1] = res.p[0], res must be clockwise
double dia_rotating_calipers(Poly &res, int &ix, int &jx) {
double dia = ;
int q = ;
for(int i = ; i < res.n - ; ++i) {
while(sgn(cross(res.p[i], res.p[q + ], res.p[i + ]) - cross(res.p[i], res.p[q], res.p[i + ])) > )
q = (q + ) % (res.n - );
if(sgn(dist(res.p[i], res.p[q]) - dia) > ) {
dia = dist(res.p[i], res.p[q]);
ix = i; jx = q;
}
if(sgn(dist(res.p[i + ], res.p[q]) - dia) > ) {
dia = dist(res.p[i + ], res.p[q]);
ix = i + ; jx = q;
}
}
return dia;
}
//a and b must be clockwise, find the minimum distance between two convex hull
double half_rotating_calipers(Poly &a, Poly &b) {
int sa = , sb = ;
for(int i = ; i < a.n; ++i) if(sgn(a.p[i].y - a.p[sa].y) < ) sa = i;
for(int i = ; i < b.n; ++i) if(sgn(b.p[i].y - b.p[sb].y) < ) sb = i;
double tmp, ans = dist(a.p[], b.p[]);
for(int i = ; i < a.n; ++i) {
while(sgn(tmp = cross(a.p[sa], a.p[sa + ], b.p[sb + ]) - cross(a.p[sa], a.p[sa + ], b.p[sb])) > )
sb = (sb + ) % (b.n - );
if(sgn(tmp) < ) ans = min(ans, Point_to_Seg(b.p[sb], Seg(a.p[sa], a.p[sa + ])));
else ans = min(ans, Seg_to_Seg(Seg(a.p[sa], a.p[sa + ]), Seg(b.p[sb], b.p[sb + ])));
sa = (sa + ) % (a.n - );
}
return ans;
} double rotating_calipers(Poly &a, Poly &b) {
return min(half_rotating_calipers(a, b), half_rotating_calipers(b, a));
} /*******************************************************************************************/ Point p[MAXN];
Circle cir;
double r;
int n; int main() {
while(scanf("%lf", &r) != EOF) {
scanf("%d", &n);
for(int i = ; i < n; ++i) p[i].read();
p[n] = p[];
cir = Circle(Point(, ), r);
double ans = ;
for(int i = ; i < n; ++i)
ans += sgn(cross(p[i], Point(, ), p[i + ])) * CommonArea(cir, p[i], p[i + ]);
printf("%.2f\n", fabs(ans));
}
}
POJ 3675 Telescope(简单多边形和圆的面积交)的更多相关文章
- POJ 3675 Telescope 简单多边形和圆的面积交
这道题得控制好精度,不然会贡献WA QAQ 还是那个规则: int sgn(double x){ if(x > eps) return 1; else if(x < - eps) ret ...
- POJ 3675 Telescope
题意:给定一个不自交的多边形,要求和圆心在原点的圆的面积交. 思路:同POJ2986,是加强版 代码: #include<algorithm> #include<cstdio> ...
- poj 3675 Telescope (圆与多边形面积交)
3675 -- Telescope 再来一题.这题的代码还是继续完全不看模板重写的. 题意不解释了,反正就是一个单纯的圆与多边形的交面积. 这题的精度有点搞笑.我用比较高的精度来统计面积,居然wa了. ...
- poj3675 求多边形与圆的面积交
题意:给出多边形的顶点坐标.圆的圆心坐标和半径,求面积交 sol:又是模板题啦= = 注意poj的C++好像认不出hypot函数,要稍微改写一下. hypot(double x,double y):即 ...
- hdu 5120 Intersection 两个圆的面积交
Intersection Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others) P ...
- HDU 5120 Intersection (圆的面积交)
题意:给定两个圆环,求两个圆环的面积交. 析:很容易知道,圆环面积交就是,大圆与大圆面积交 - 大圆和小圆面积交 - 小圆和大圆面积交 + 小圆和小圆面积交. 代码如下: #pragma commen ...
- HDU - 2892:area (圆与多边形交 求面积)
pro:飞行员去轰炸一个小岛,给出炸弹落地点的位置信息,以及轰炸半径:按顺时针或者逆时针给出小岛的边界点. 求被轰炸的小岛面积. sol:即是求圆和多边形的面积交. (只会套板子的我改头换面,先理解然 ...
- Harry Potter and J.K.Rowling(半平面交+圆和矩形交)
Harry Potter and J.K.Rowling http://acm.hdu.edu.cn/showproblem.php?pid=3982 Time Limit: 2000/1000 MS ...
- poj 2546(两圆公共面积)
Circular Area Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5682 Accepted: 2225 Des ...
随机推荐
- chromium之lazy_instance
先看看介绍 // The LazyInstance<Type, Traits> class manages a single instance of Type, // which will ...
- Maven DebugConfiguration配置运行内存
-Xms256M -Xmx512M -XX:PermSize=256m -XX:MaxPermSize=512m
- vue调用豆瓣API加载图片403问题
"豆瓣API是有请求次数限制的”,这会引发图片在加载的时候出现403问题,视图表现为“图片加载不出来”,控制台表现为报错403. 其实是豆瓣限制了图片的加载,我自己用了一个办法把图片缓存下来 ...
- h5跳转到app的实现
随着业务的增加,可能存在这么一种需求,就是需要从h5中直接跳转到app.如果没有安装app的话,则提示到应用市场或者app store下载安装.不过问题就在这个地方,单纯的用h5是没有方法判断是否安装 ...
- day 22 内置的模块
1.简单的理解模块 写的每一个py文件都是一个模块,还有一些是我们一直在使用的模块. buildins: 内置模块. print, input random 主要是和随机相关的内容: ...
- IDELPHI是一个MIS系统初学者的乐园空间
DELPHI的长处之一是MIS系统,此空间介绍了MIS中种种问题,及使用DELPHI XE做的办法. 为什么选择DELPHI? DELPHI是开发效率很高的一个工具,但也许很多人都喜欢选择微软.NET ...
- 用Python生成词云
词云以词语为基本单元,根据词语在文本中出现的频率设计不同大小的形状以形成视觉上的不同效果,从而使读者只要“一瞥“即可领略文本的主旨.以下是一个词云的简单示例: import jieba from wo ...
- linux shell 字符串常用操作
1.shell内置的字符串操作 表达式 含义 ${#string} $string的长度 ${string:position} string中,从位置$position开始提取字符串 ${string ...
- 批量改变图片的尺寸大小 python opencv
我目标文件夹下有一大批图片,我要把它转变为指定尺寸大小的图片,用pthon和opencv实现的. 以上为原图片. import cv2 import os # 按指定图像大小调整尺寸 def resi ...
- 关于条件约束问题的无偏差统计——一个偏差控制型生成器(Unbiased Statistics of a Constraint Satisfaction Problem – a Controlled-Bias Generator——by Denis Berthier)
论文地址:https://hal.archives-ouvertes.fr/hal-00641955 Unbiased Statistics of a Constraint Satisfaction ...