poj 1584 A Round Peg in a Ground Hole 多边形凸性判断，点与多边形位置关系，圆与多边形位置关系

题目地址：poj1584

额 全部是模板，凸性判断只需要“一直绕着一个方向转”。 不需要求“最多定点的凸包”然后判断点数是否相等。

不是凸多边形 就输出HOLE IS ILL-FORMED（注意只要不出现符号完全冲突--有0就不算冲突） 就是凸~

凸多边形，而且在多变形内而且圆的半径最大是相切--才输出fit

其余 not fit

代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#include<string>

const  double INF=0x3fffffff;

const  double eps=1e-10;
const double PI=acos(-1.0);

using namespace std;

struct Point{
double x;
double y;
Point(double x=0,double y=0):x(x),y(y){}

};

int dcmp(double x)  {return (x>eps)-(x<-eps); }
int sgn(double x)  {return (x>eps)-(x<-eps); }
typedef  Point  Vector;

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);}

ostream &operator<<(ostream & out,Point & P) { out<<P.x<<' '<<P.y; return out;}
//
bool  operator< (const Point &A,const Point &B) { return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0); }

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  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.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  Area2(Point A,Point B,Point C ) {return Cross(B-A, C-A);}

Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}
Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}

Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)   //    v,w 不平行！
{
Vector u=P-Q;
double t=Cross(w, u)/Cross(v,w);
return P+v*t;

}

double DistanceToLine(Point P,Point A,Point B)    // A,B 不重合则完全没问题
{
Vector v1=P-A; Vector v2=B-A;
return fabs(Cross(v1,v2))/Length(v2);

}

double DistanceToSegment(Point P,Point A,Point B)
{
if(A==B)  return Length(P-A);

Vector v1=B-A;
Vector v2=P-A;
Vector v3=P-B;

if(dcmp(Dot(v1,v2))==-1)    return  Length(v2);
else if(Dot(v1,v3)>0)    return Length(v3);

else return DistanceToLine(P, A, B);

}

Point GetLineProjection(Point P,Point A,Point B)      //  A,B 不重合就行
{
Vector v=B-A;
Vector v1=P-A;
double t=Dot(v,v1)/Dot(v,v);

return  A+v*t;
}

bool  SegmentProperIntersection(Point a1,Point a2,Point b1,Point b2)
{
double c1=Cross(b1-a1, a2-a1);
double c2=Cross(b2-a1, a2-a1);
double c3=Cross(a1-b1, b2-b1);
double c4=Cross(a2-b1, b2-b1);

return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0 ;

}

bool  OnSegment(Point P,Point A,Point B)
{
return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;
}

bool  SegmentCommon(Point a1,Point a2,Point b1,Point b2)
{
double c1=Cross(b1-a1, a2-a1);
double c2=Cross(b2-a1, a2-a1);
double c3=Cross(a1-b1, b2-b1);
double c4=Cross(a2-b1, b2-b1);

if(c1==0&&c2==0&&c3==0&&c4==0)
{
if(OnSegment(a1, b1, b2))  return 1;
if(OnSegment(a2, b1, b2))  return 1;

return 0;
}

return dcmp(c1)*dcmp(c2)<=0&&dcmp(c3)*dcmp(c4)<=0 ;

}

double PolygonArea(Point *p,int n)
{
double area=0;

for(int i=1;i<n-1;i++)
{
area+=Cross(p[i]-p[0], p[i+1]-p[0]);

}
return area/2;

}

Point  read_point()
{
Point P;

scanf("%lf%lf",&P.x,&P.y);

return  P;
}

// ---------------与圆有关的--------

struct Circle
{
Point c;
double r;

Circle(Point c=Point(0,0),double r=0):c(c),r(r) {}

Point point(double a)
{
return Point(c.x+r*cos(a),c.y+r*sin(a));
}

};

struct  Line
{
Point p;
Vector v;
Line(Point p=Point(0,0),Vector v=Vector(0,1)):p(p),v(v) {}

Point point(double t)
{
return Point(p+v*t);
}

};

int getLineCircleIntersection(Line L,Circle C,double &t1,double &t2,vector<Point> &sol)
{
double a=L.v.x;
double b=L.p.x-C.c.x;
double c=L.v.y;
double d=L.p.y-C.c.y;

double e=a*a+c*c;
double f=2*(a*b+c*d);
double g=b*b+d*d-C.r*C.r;

double delta=f*f-4*e*g;

if(dcmp(delta)<0) return 0;

if(dcmp(delta)==0)
{
t1=t2=-f/(2*e);
sol.push_back(L.point(t1));
return 1;
}

else
{
t1=(-f-sqrt(delta))/(2*e);
t2=(-f+sqrt(delta))/(2*e);

sol.push_back(L.point(t1));
sol.push_back(L.point(t2));

return 2;
}

}

// 向量极角公式

double angle(Vector v)  {return atan2(v.y,v.x);}

int getCircleCircleIntersection(Circle C1,Circle C2,vector<Point> &sol)
{
double d=Length(C1.c-C2.c);

if(dcmp(d)==0)
{
if(dcmp(C1.r-C2.r)==0)  return -1;  // 重合
else return 0;    //  内含  0 个公共点
}

if(dcmp(C1.r+C2.r-d)<0)  return 0;  // 外离
if(dcmp(fabs(C1.r-C2.r)-d)>0)  return 0;  // 内含

double a=angle(C2.c-C1.c);
double da=acos((C1.r*C1.r+d*d-C2.r*C2.r)/(2*C1.r*d));

Point p1=C1.point(a-da);
Point p2=C1.point(a+da);

sol.push_back(p1);

if(p1==p2)  return 1; // 相切
else
{
sol.push_back(p2);
return 2;
}
}

//  求点到圆的切线

int getTangents(Point p,Circle C,Vector *v)
{
Vector u=C.c-p;

double dist=Length(u);

if(dcmp(dist-C.r)<0)  return 0;

else if(dcmp(dist-C.r)==0)
{
v[0]=Rotate(u,PI/2);
return 1;
}

else
{

double ang=asin(C.r/dist);
v[0]=Rotate(u,-ang);
v[1]=Rotate(u,+ang);
return 2;
}

}

//  求两圆公切线

int getTangents(Circle A,Circle B,Point *a,Point *b)
{
int cnt=0;

if(A.r<B.r)
{
swap(A,B); swap(a, b);  //  有时需标记
}

double d=Length(A.c-B.c);

double rdiff=A.r-B.r;
double rsum=A.r+B.r;

if(dcmp(d-rdiff)<0)  return 0;   // 内含

double base=angle(B.c-A.c);

if(dcmp(d)==0&&dcmp(rdiff)==0)   return -1 ;  // 重合 无穷多条切线

if(dcmp(d-rdiff)==0)             // 内切   外公切线
{
a[cnt]=A.point(base);
b[cnt]=B.point(base);
cnt++;
return 1;
}

// 有外公切线的情形

double ang=acos(rdiff/d);
a[cnt]=A.point(base+ang);
b[cnt]=B.point(base+ang);
cnt++;
a[cnt]=A.point(base-ang);
b[cnt]=B.point(base-ang);
cnt++;

if(dcmp(d-rsum)==0)     // 外切 有内公切线
{
a[cnt]=A.point(base);
b[cnt]=B.point(base+PI);
cnt++;
}

else  if(dcmp(d-rsum)>0)   // 外离   又有两条外公切线
{
double  ang_in=acos(rsum/d);
a[cnt]=A.point(base+ang_in);
b[cnt]=B.point(base+ang_in+PI);
cnt++;
a[cnt]=A.point(base-ang_in);
b[cnt]=B.point(base-ang_in+PI);
cnt++;
}

return cnt;
}

//  几何算法模板

int  isPointInPolygon(Point p,Point * poly,int n)
{
int wn=0;
for(int i=0;i<n;i++)
{
if(OnSegment(p, poly[i], poly[(i+1)%n]))  return -1;
int k=dcmp(Cross(poly[(i+1)%n]-poly[i], p-poly[i]));
int d1=dcmp(poly[i].y-p.y);
int d2=dcmp(poly[(i+1)%n].y-p.y);

if(k>0&&d1<=0&&d2>0) wn++;
if(k<0&&d2<=0&&d1>0) wn--;

}

if(wn!=0)  return 1;
else   return 0;

}

int ConvexHull(Point *p,int n,Point *ch)
{
sort(p,p+n);
int m=0;
for(int i=0;i<n;i++)
{
while(m>1&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<0) m--;
ch[m++]=p[i];
}

int k=m;
for(int i=n-1;i>=0;i--)
{
while(m>k&&dcmp(Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2]))<0) m--;
ch[m++]=p[i];
}

if(n>1)  m--;

return m;
}

int n;
Point p[100000];  //  题目连n的范围都没给...
Point ch[100000];

int main()
{
Circle C;
while(cin>>n)
{
if(n<3)  break;
scanf("%lf",&C.r);
C.c=read_point();
for(int i=0;i<n;i++)
{
p[i]=read_point();

}

p
=p[0];

int sg=0;
bool first=1;
Vector v=p[1]-p[0];

bool isConvex=1;

for(int i=1;i<n;i++)
{
if(first) sg=dcmp(Cross(v, p[i+1]-p[i]));
if(first) first=0;
if(dcmp(Cross( p[i]-p[i-1],p[i+1]-p[i]))<0&&sg>=0)
{
isConvex=0;
break;
}
if(dcmp(Cross( p[i]-p[i-1],p[i+1]-p[i]))>=0&&sg<0)
{
isConvex=0;
break;
}
}

if(!isConvex)
{
cout<<"HOLE IS ILL-FORMED"<<endl;
}

else
{

bool inside=isPointInPolygon(C.c, p, n);

if(inside)
{
bool match=1;
for(int i=0;i<n;i++)
{
if(dcmp(DistanceToLine(C.c, p[i], p[i+1])-C.r)<0)
{
match=0;
break;
}
}

if(match)   cout<<"PEG WILL FIT"<<endl;
else cout<<"PEG WILL NOT FIT"<<endl;

}
else
{
cout<<"PEG WILL NOT FIT"<<endl;
}

}

}

}