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; }

View Code

 

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;

}

View Code