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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:

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#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#define pi acos(-1.0)#define eps 1e-8using 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;}
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参考文章: http://blog.csdn.net/lyy289065406/article/details/6648606

POJ 1584 A Round Peg in a Ground Hole --判定点在形内形外形上