HDU 5230 (计算几何 圆和多边形面积交)

题目链接:点击这里

题意:给出一个多边形区域, 给出两个点 A , B , 某一个点距离 B 的距离为 dB , 距离 A 的距离为 dA , 如果 k×dA≤dB 一个点是安全的. 求安全的面积.

先假设两个点是 (xA,yA) , (xB,yB) , 假设点 (x,y) 是安全的, 那么满足 d×(x−xA)2+(y−ya)2−−−−−−−−−−−−−−−−−√≤(x−xB)2+(y−yB)2−−−−−−−−−−−−−−−−−−√

平方整理一下就是 (d2−1)x2−(2dxA−2xB)x+d2x2A−x2B+(d2−1)y−(2dyA−2yB)y+d2y2A−y2B≤0

这个方程整理一下就是圆方程了. 所以所求的面积就是一个圆和一个多边形的面积交.

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

const double eps = 1e-8;
const double INF = 1e20;
const double pi = acos (-1.0);

int dcmp (double x) {
if (fabs (x) < eps) return 0;
return (x < 0 ? -1 : 1);
}
inline double sqr (double x) {return x*x;}

//*************点
struct Point {
    double x, y;
    Point (double _x = 0, double _y = 0):x(_x), y(_y) {}
    void input () {scanf ("%lf%lf", &x, &y);}
    void output () {printf ("%.2f %.2f\n", x, y);}
    bool operator == (const Point &b) const {
return (dcmp (x-b.x) == 0 && dcmp (y-b.y) == 0);
    }
    bool operator < (const Point &b) const {
return (dcmp (x-b.x) == 0 ? dcmp (y-b.y) < 0 : x < b.x);
    }
    Point operator + (const Point &b) const {
return Point (x+b.x, y+b.y);
    }
    Point operator - (const Point &b) const {
return Point (x-b.x, y-b.y);
    }
    Point operator * (double a) {
return Point (x*a, y*a);
    }
    Point operator / (double a) {
return Point (x/a, y/a);
 }
 double len2 () {//返回长度的平方
return sqr (x) + sqr (y);
    }
    double len () {//返回长度
return sqrt (len2 ());
    }
    Point change_len (double r) {//转化为长度为r的向量
        double l = len ();
if (dcmp (l) == 0) return *this;//零向量返回自身
        r /= l;
return Point (x*r, y*r);
    }
    Point rotate_left () {//顺时针旋转90度
return Point (-y, x);
    }
    Point rotate_right () {//逆时针旋转90度
return Point (y, -x);
    }
    Point rotate (Point p, double ang) {//绕点p逆时针旋转ang
        Point v = (*this)-p;
        double c = cos (ang), s = sin (ang);
return Point (p.x + v.x*c - v.y*s, p.y + v.x*s + v.y*c);
    }
    Point normal () {//单位法向量
        double l = len ();
return Point (-y/l, x/l);
 }
};

double cross (Point a, Point b) {//叉积

 return a.x*b.y-a.y*b.x;
}
double dot (Point a, Point b) {//点积

 return a.x*b.x + a.y*b.y;
}
double dis (Point a, Point b) {//两个点的距离

 Point p = b-a; return p.len ();
}
double rad_degree (double rad) {//弧度转化为角度

 return rad/pi*180;
}
double rad (Point a, Point b) {//两个向量的夹角

return fabs (atan2 (fabs (cross (a, b)), dot (a, b)) );
}
bool parallel (Point a, Point b) {//向量平行
    double p = rad (a, b);
return dcmp (p) == 0 || dcmp (p-pi) == 0;
}

//************直线 线段
struct Line {
    Point s, e;//直线的两个点
    Line () {}
    Line (Point _s, Point _e) : s(_s), e(_e) {}
    //一个点和倾斜角确定直线
    Line (Point p, double ang) {
s = p;
if (dcmp (ang-pi/2) == 0) {
            e = s + Point (0, 1);
        }
else
            e = s + Point (1, tan (ang));
    }
    //ax+by+c=0确定直线
    Line (double a, double b, double c) {
if (dcmp (a) == 0) {
s = Point (0, -c/b);
            e = Point (1, -c/b);
        }
else if (dcmp (b) == 0) {
s = Point (-c/a, 0);
            e = Point (-c/a, 1);
        }
else {
s = Point (0, -c/b);
            e = Point (1, (-c-a)/b);
        }
    }
    void input () {
s.input ();
        e.input ();
    }
    void adjust () {
if (e < s) swap (e, s);
    }
    double length () {//求线段长度
return dis (s, e);
    }
    double angle () {//直线的倾斜角
        double k = atan2 (e.y-s.y, e.x-s.x);
if (dcmp (k) < 0) k += pi;
if (dcmp (k-pi) == 0) k -= pi;
return k;
    }
};

double point_to_line (Point p, Line a) {//点到直线的距离
return fabs (cross (p-a.s, a.e-a.s) / a.length ());
}

double point_to_seg (Point p, Line a) {//点到线段的距离
if (dcmp (dot (p-a.s, a.e-a.s)) < 0 || dcmp (dot (p-a.e, a.s-a.e)) < 0)
return min (dis (p, a.e), dis (p, a.s));
return point_to_line (p, a);
}

Point projection (Point p, Line a) {//点在直线上的投影
return a.s + (((a.e-a.s) * dot (a.e-a.s, p-a.s)) / (a.e-a.s).len2() );
}
//***************圆
struct Circle {
//圆心 半径
    Point p;
    double r;
    Circle () {}
    Circle (Point _p, double _r) : p(_p), r(_r) {}
    Circle (double a, double b, double _r) {
        p = Point (a, b);
        r = _r;
    }
    void input () {
        p.input ();
        scanf ("%lf", &r);
    }
    void output () {
        p.output ();
printf (" %.2f\n", r);
    }
    bool operator == (const Circle &a) const {
return p == a.p && (dcmp (r-a.r) == 0);
    }
    double area () {//面积
return pi*r*r;
    }
    double circumference () {//周长
return 2*pi*r;
    }
    bool operator < (const Circle &a) const {
return p < a.p || (p == a.p && r < a.r);
    }
};

int relation (Point p, Circle a) {//点和圆的关系
    //0:圆外 1:圆上 2:圆内
    double d = dis (p, a.p);
if (dcmp (d-a.r) == 0) return 1;
return (dcmp (d-a.r) < 0 ? 2 : 0);
}

int relation (Line a, Circle b) {//直线和圆的关系
    //0:相离 1:相切 2:相交
    double p = point_to_line (b.p, a);
if (dcmp (p-b.r) == 0) return 1;
return (dcmp (p-b.r) < 0 ? 2 : 0);
}

int relation (Circle a, Circle v) {//两圆的位置关系
    //1:内含 2:内切 3:相交 4:外切 5:相离
    double d = dis (a.p, v.p);
if (dcmp (d-a.r-v.r) > 0) return 5;
if (dcmp (d-a.r-v.r) == 0) return 4;
    double l = fabs (a.r-v.r);
if (dcmp (d-a.r-v.r) < 0 && dcmp (d-l) > 0) return 3;
if (dcmp (d-l) == 0) return 2;
if (dcmp (d-l) < 0) return 1;
}

int circle_intersection (Circle a, Circle v, Point &p1, Point &p2) {//两个圆的交点
    //返回交点个数 交点保存在引用中
int rel = relation (a, v);
if (rel == 1 || rel == 5) return 0;
    double d = dis (a.p, v.p);
    double l = (d*d + a.r*a.r - v.r*v.r) / (2*d);
    double h = sqrt (a.r*a.
                     r - l*l);
    Point tmp = a.p + (v.p-a.p).change_len (l);
    p1 = tmp + ((v.p-a.p).rotate_left ().change_len (h));
    p2 = tmp + ((v.p-a.p).rotate_right ().change_len (h));
if (rel == 2 || rel == 4) return 1;
return 2;
}

int line_cirlce_intersection (Line v, Circle u, Point &p1, Point &p2) {//直线和圆的交点
    //返回交点个数 交点保存在引用中
if (!relation (v, u)) return 0;
    Point a = projection (u.p, v);
    double d = point_to_line (u.p, v);
    d = sqrt (u.r*u.r - d*d);
if (dcmp (d) == 0) {
        p1 = a, p2 = a;
return 1;
    }
    p1 = a + (v.e-v.s).change_len (d);
    p2 = a - (v.e-v.s).change_len (d);
return 2;
}

double circle_traingle_area (Point a, Point b, Circle c) {//圆心三角形的面积
    //a.output (), b.output (), c.output ();
    Point p = c.p; double r = c.r; //cout << cross (p-a, p-b) << endl;
if (dcmp (cross (p-a, p-b)) == 0) return 0;
    Point q[5];
int len = 0;
q[len++] = a;
    Line l(a, b);
    Point p1, p2;
if (line_cirlce_intersection (l, c, q[1], q[2]) == 2) {
if (dcmp (dot (a-q[1], b-q[1])) < 0) q[len++] = q[1];
if (dcmp (dot (a-q[2], b-q[2])) < 0) q[len++] = q[2];
    }
q[len++] = b;
if (len == 4 && dcmp (dot (q[0]-q[1], q[2]-q[1])) > 0)
        swap (q[1], q[2]);
    double res = 0;
for (int i = 0; i < len-1; i++) {
if (relation (q[i], c) == 0 || relation (q[i+1], c) == 0) {
            double arg = rad (q[i]-p, q[i+1]-p);
            res += r*r*arg/2.0;
        }
else {
            res += fabs (cross (q[i]-p, q[i+1]-p))/2;
        }
    }
return res;
}
double area_polygon_circle (Circle c, Point *p, int n) {//多边形和圆交面积
    double ans = 0;
for (int i = 0; i < n; i++) {
int j = (i+1)%n;
if (dcmp (cross (p[j]-c.p, p[i]-c.p)) >= 0)
            ans += circle_traingle_area (p[i], p[j], c);
else
            ans -= circle_traingle_area (p[i], p[j], c);
    }
return fabs (ans);
}

#define maxn 511
Point p[maxn];
int n;
double R, k;
Point A, B;

void solve () {
    double aa, bb, pp, qq;
    double gg = 1-k*k;
    aa = (2*B.x-2*A.x*k*k)/gg;
    bb = (B.x*B.x-k*k*A.x*A.x)/gg;
    pp = (2*B.y-2*A.y*k*k)/gg;
    qq = (B.y*B.y-k*k*A.y*A.y)/gg;
    Point O (aa/2, pp/2);
    double R = sqrt (-bb+aa*aa/4 - qq+pp*pp/4);
    Circle C (O, R);
    double ans = area_polygon_circle (C, p, n);
printf("%.10f\n",ans);
}

int main() {
int kase= 0 ;
while(scanf ("%d%lf", &n, &k) == 2) {
printf ("Case %d: ", ++kase);
for (int i = 0; i < n; i++) {
            p[i].input ();
        }
        A.input (), B.input ();
        solve ();
    }
return 0;
}