题目链接:uva 10321 - Polygon Intersection
就是两个多边形线段的交点,并且判断出所有在多边形内部的点。
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const double pi = 4 * atan(1);
const double eps = 1e-9;
inline int dcmp (double x) { if (fabs(x) < eps) return 0; else return x < 0 ? -1 : 1; }
inline double getDistance (double x, double y) { return sqrt(x * x + y * y); }
struct Point {
double x, y;
Point (double x = 0, double y = 0): x(x), y(y) {}
void read () { scanf("%lf%lf", &x, &y); }
void write () { printf("%lf %lf", x, y); }
bool operator == (const Point& u) const { return dcmp(x - u.x) == 0 && dcmp(y - u.y) == 0; }
bool operator != (const Point& u) const { return !(*this == u); }
bool operator < (const Point& u) const { return x < u.x || (x == u.x && y < u.y); }
bool operator > (const Point& u) const { return u < *this; }
bool operator <= (const Point& u) const { return *this < u || *this == u; }
bool operator >= (const Point& u) const { return *this > u || *this == u; }
Point operator + (const Point& u) { return Point(x + u.x, y + u.y); }
Point operator - (const Point& u) { return Point(x - u.x, y - u.y); }
Point operator * (const double u) { return Point(x * u, y * u); }
Point operator / (const double u) { return Point(x / u, y / u); }
double operator * (const Point& u) { return x*u.y - y*u.x; }
};
typedef Point Vector;
struct Line {
double a, b, c;
Line (double a = 0, double b = 0, double c = 0): a(a), b(b), c(c) {}
};
struct Circle {
Point o;
double r;
Circle () {}
Circle (Point o, double r = 0): o(o), r(r) {}
void read () { o.read(), scanf("%lf", &r); }
Point point(double rad) { return Point(o.x + cos(rad)*r, o.y + sin(rad)*r); }
double getArea (double rad) { return rad * r * r / 2; }
};
namespace Punctual {
double getDistance (Point a, Point b) { double x=a.x-b.x, y=a.y-b.y; return sqrt(x*x + y*y); }
};
namespace Vectorial {
/* 点积: 两向量长度的乘积再乘上它们夹角的余弦, 夹角大于90度时点积为负 */
double getDot (Vector a, Vector b) { return a.x * b.x + a.y * b.y; }
/* 叉积: 叉积等于两向量组成的三角形有向面积的两倍, cross(v, w) =