POJ 1584 A Round Peg in a Ground Hole --判定点在形内形外形上
题意: 给一个圆和一个多边形,多边形点可能按顺时针给出,也可能按逆时针给出,先判断多边形是否为凸包,再判断圆是否在凸包内。
解法: 先判是否为凸包,沿着i=0~n,先得出初始方向dir,dir=1为逆时针,dir=-1为顺时针,然后如果后面有两个相邻的边叉积后得出旋转方向为nowdir,如果dir*nowdir < 0,说明方向逆转了,即出现了凹点,说明不是凸多边形。
然后判圆是否在多边形内: 先判圆心是否在多边形内,用环顾法,然后如果在之内,则依次判断圆心与每条凸包边的距离与半径的距离,如果所有的dis都大于等于R,说明圆在凸包内。
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#define pi acos(-1.0)
#define eps 1e-8
using namespace std;
struct Point{
double x,y;
Point(double x=0, double y=0):x(x),y(y) {}
void input() { scanf("%lf%lf",&x,&y); }
};
typedef Point Vector;
struct Circle{
Point c;
double r;
Circle(){}
Circle(Point c,double r):c(c),r(r) {}
Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); }
void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); }
};
int dcmp(double x) {
if(x < -eps) return -1;
if(x > eps) return 1;
return 0;
}
template <class T> T sqr(T x) { return x * x;}
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); }
Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); }
Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); }
bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }
bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; }
bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; }
bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; }
double Length(Vector A) { return sqrt(Dot(A, A)); }
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); }
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; }
double DistanceToSeg(Point P, Point A, Point B)
{
if(A == B) return Length(P-A);
Vector v1 = B-A, v2 = P-A, v3 = P-B;
if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
return fabs(Cross(v1, v2)) / Length(v1);
}
//点是否在多边形内部
int CheckPointInPolygon(Point A,Point* p,int n){
double TotalAngle = 0.0;
for(int i=0;i<n;i++) {
if(dcmp(Cross(p[i]-A,p[(i+1)%n]-A)) >= 0) TotalAngle += Angle(p[i]-A,p[(i+1)%n]-A);
else TotalAngle -= Angle(p[i]-A,p[(i+1)%n]-A);
}
if(dcmp(TotalAngle) == 0) return 0; //外部
else if(dcmp(fabs(TotalAngle)-2*pi) == 0) return 1; //完全内部
else if(dcmp(fabs(TotalAngle)-pi) == 0) return 2; //边界上
else return 3; //多边形顶点
}
//判断未知时针方向的多边形是否是凸包
bool CheckConvexHull(Point* p,int n){
int dir = 0; //旋转方向
for(int i=0;i<n;i++) {
int nowdir = dcmp(Cross(p[(i+1)%n]-p[i],p[(i+2)%n]-p[i]));
if(!dir) dir = nowdir;
if(dir*nowdir < 0) return false; //非凸包
}
return true;
}
Point p[107];
int main()
{
int n,i,j;
Circle Peg;
while(scanf("%d",&n)!=EOF && n >= 3)
{
scanf("%lf",&Peg.r); Peg.c.input();
for(i=0;i<n;i++) p[i].input();
if(!CheckConvexHull(p,n)) { puts("HOLE IS ILL-FORMED"); continue; }
if(CheckPointInPolygon(Peg.c,p,n))
{
for(i=0;i<n;i++)
{
double dis = DistanceToSeg(Peg.c,p[i],p[(i+1)%n]);
if(dcmp(dis-Peg.r) < 0) break;
}
if(i == n) { puts("PEG WILL FIT"); continue; }
}
puts("PEG WILL NOT FIT");
}
return 0;
}
参考文章: http://blog.csdn.net/lyy289065406/article/details/6648606
射线法:
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> #define eps 1e-8 using namespace std; struct Point{ double x,y; Point(double x=0, double y=0):x(x),y(y) {} void input() { scanf("%lf%lf",&x,&y); } }; typedef Point Vector; struct Circle{ Point c; double r; Circle(){} Circle(Point c,double r):c(c),r(r) {} Point point(double a) { return Point(c.x + cos(a)*r, c.y + sin(a)*r); } void input() { scanf("%lf%lf%lf",&c.x,&c.y,&r); } }; int dcmp(double x) { if(x < -eps) return -1; if(x > eps) return 1; return 0; } template <class T> T sqr(T x) { return x * x;} Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); } Vector operator - (Vector A, Vector B) { return Vector(A.x - B.x, A.y - B.y); } Vector operator * (Vector A, double p) { return Vector(A.x*p, A.y*p); } Vector operator / (Vector A, double p) { return Vector(A.x/p, A.y/p); } bool operator < (const Point& a, const Point& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); } bool operator >= (const Point& a, const Point& b) { return a.x >= b.x && a.y >= b.y; } bool operator <= (const Point& a, const Point& b) { return a.x <= b.x && a.y <= b.y; } bool operator == (const Point& a, const Point& b) { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; } double Dot(Vector A, Vector B) { return A.x*B.x + A.y*B.y; } double Length(Vector A) { return sqrt(Dot(A, A)); } double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } Vector VectorUnit(Vector x){ return x / Length(x);} Vector Normal(Vector x) { return Point(-x.y, x.x) / Length(x);} double angle(Vector v) { return atan2(v.y, v.x); } bool OnSegment(Point P, Point A, Point B) { return dcmp(Cross(A-P,B-P)) == 0 && dcmp(Dot(A-P,B-P)) <= 0; } double DistanceToSeg(Point P, Point A, Point B) { if(A == B) return Length(P-A); Vector v1 = B-A, v2 = P-A, v3 = P-B; if(dcmp(Dot(v1, v2)) < 0) return Length(v2); if(dcmp(Dot(v1, v3)) > 0) return Length(v3); return fabs(Cross(v1, v2)) / Length(v1); } //判断未知时针方向的多边形是否是凸包 bool CheckConvexHull(Point* p,int n){ int dir = 0; //旋转方向 for(int i=0;i<n;i++) { int nowdir = dcmp(Cross(p[(i+1)%n]-p[i],p[(i+2)%n]-p[i])); if(!dir) dir = nowdir; if(dir*nowdir < 0) return false; //非凸包 } return true; } int Ray_PointInPolygon(Point A,Point* p,int n) { int wn = 0; for(int i=0;i<n;i++) { //if(OnSegment(A,p[i],p[(i+1)%n])) return -1; //边界 int k = dcmp(Cross(p[(i+1)%n]-p[i], A-p[i])); int d1 = dcmp(p[i].y-A.y); int d2 = dcmp(p[(i+1)%n].y-A.y); if(k > 0 && d1 <= 0 && d2 > 0) wn++; if(k < 0 && d2 <= 0 && d1 > 0) wn--; } if(wn) return 1; //内部 return 0; //外部 } Point p[107]; int main() { int n,i,j; Circle Peg; while(scanf("%d",&n)!=EOF && n >= 3) { scanf("%lf",&Peg.r); Peg.c.input(); for(i=0;i<n;i++) p[i].input(); if(!CheckConvexHull(p,n)) { puts("HOLE IS ILL-FORMED"); continue; } if(Ray_PointInPolygon(Peg.c,p,n)) { for(i=0;i<n;i++) { double dis = DistanceToSeg(Peg.c,p[i],p[(i+1)%n]); if(dcmp(dis-Peg.r) < 0) break; } if(i == n) { puts("PEG WILL FIT"); continue; } } puts("PEG WILL NOT FIT"); } return 0; }
作者:whatbeg
出处1:http://whatbeg.com/
出处2:http://www.cnblogs.com/whatbeg/
本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利。
更多精彩文章抢先看?详见我的独立博客: whatbeg.com