2014
02-17

# area

0 0 2000
100 0
100

4
1900 100
2000 100
2000 -100
1900 -100

15707.96

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
const int maxn = 111111;
const int maxisn = 21;
const double eps = 1e-8;
const double pi = acos(-1.0);
int dcmp(double x)
{
if(x > eps) return 1;
return x < -eps ? -1 : 0;
}
struct Point
{
double x, y;
Point(){x = y = 0;}
Point(double a, double b)
{x = a, y = b;}
inline Point operator-(const Point &b)const
{return Point(x - b.x, y - b.y);}
inline Point operator+(const Point &b)const
{return Point(x + b.x, y + b.y);}
inline Point operator*(const double &b)const
{return Point(x * b, y * b);}
inline double dot(const Point &b)const
{return x * b.x + y * b.y;}
inline double cross(const Point &b, const Point &c)const
{return (b.x - x) * (c.y - y) - (c.x - x) * (b.y - y);}
inline double Dis(const Point &b)const
{return sqrt((*this - b).dot(*this - b));}
inline bool InLine(const Point &b, const Point &c)const//三点共线
{return !dcmp(cross(b, c));}
inline bool OnSeg(const Point &b, const Point &c)const//点在线段上，包括端点
{return InLine(b, c) && (*this - c).dot(*this - b) < eps;}
};
inline double min(double a, double b)
{return a < b ? a : b;}
inline double max(double a, double b)
{return a > b ? a : b;}
inline double Sqr(double x)
{return x * x;}
inline double Sqr(const Point &p)
{return p.dot(p);}
Point LineCross(const Point &a, const Point &b, const Point &c, const Point &d)
{
double u = a.cross(b, c), v = b.cross(a, d);
return Point((c.x * v + d.x * u) / (u + v), (c.y * v + d.y * u) / (u + v));
}
double LineCrossCircle(const Point &a, const Point &b, const Point &r,
double R, Point &p1, Point &p2)
{
Point fp = LineCross(r, Point(r.x + a.y - b.y, r.y + b.x - a.x), a, b);
double rtol = r.Dis(fp);
double rtos = fp.OnSeg(a, b) ? rtol : min(r.Dis(a), r.Dis(b));
double atob = a.Dis(b);
double fptoe = sqrt(R * R - rtol * rtol) / atob;
if(rtos > R - eps) return rtos;
p1 = fp + (a - b) * fptoe;
p2 = fp + (b - a) * fptoe;
return rtos;
}
double SectorArea(const Point &r, const Point &a, const Point &b, double R)
//不大于180度扇形面积，r->a->b逆时针
{
double A2 = Sqr(r - a), B2 = Sqr(r - b), C2 = Sqr(a - b);
return R * R * acos((A2 + B2 - C2) * 0.5 / sqrt(A2) / sqrt(B2)) * 0.5;
}
double TACIA(const Point &r, const Point &a, const Point &b, double R)
//TriangleAndCircleIntersectArea，逆时针，r为圆心
{
double adis = r.Dis(a), bdis = r.Dis(b);
if(adis < R + eps && bdis < R + eps) return r.cross(a, b) * 0.5;
Point ta, tb;
if(r.InLine(a, b)) return 0.0;
double rtos = LineCrossCircle(a, b, r, R, ta, tb);
if(rtos > R - eps) return SectorArea(r, a, b, R);
if(adis < R + eps) return r.cross(a, tb) * 0.5 + SectorArea(r, tb, b, R);
if(bdis < R + eps) return r.cross(ta, b) * 0.5 + SectorArea(r, a, ta, R);
return r.cross(ta, tb) * 0.5 +
SectorArea(r, a, ta, R) + SectorArea(r, tb, b, R);
}
double SPICA(int n, Point r, double R)//SimplePolygonIntersectCircleArea
{
int i;
Point ori, p[2];
scanf("%lf%lf", &ori.x, &ori.y);
p[0] = ori;
double res = 0, if_clock_t;
for(i = 1; i <= n; ++ i)
{
if(i == n) p[i & 1] = ori;
else scanf("%lf%lf", &p[i & 1].x, &p[i & 1].y);
if_clock_t = dcmp(r.cross(p[~i & 1], p[i & 1]));
if(if_clock_t < 0) res -= TACIA(r, p[i & 1], p[~i & 1], R);
else res += TACIA(r, p[~i & 1], p[i & 1], R);
}
return fabs(res);
}
Point boom;
int n;
double R;
int main()
{
double sx, sy, h, vx, vy;
while(scanf("%lf%lf%lf", &sx, &sy, &h) != EOF)
{
scanf("%lf%lf%lf", &vx, &vy, &R);
h = sqrt(2 * h / 10);
boom = Point(h * vx + sx, h * vy + sy);
scanf("%d", &n);
printf("%.2f\n", SPICA(n, boom, R));
}
return 0;
}

1. 学算法中的数据结构学到一定程度会乐此不疲的，比如其中的2－3树，类似的红黑树，我甚至可以自己写个逻辑文件系统结构来。