poj 1113 Wall 凸包

题目地址：poj1113

如果求出凸包，就是凸包的周长加上一整个圆的周长，证明是简单的~

代码：

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<vector>
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){}void operator<<(Point &A) {cout<<A.x<<' '<<A.y<<endl;}
};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<<endl; 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)
{Vector u=P-Q;double t=Cross(w, u)/Cross(v,w);return P+v*t;}double DistanceToLine(Point P,Point A,Point B)
{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)
{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;
}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;}//  Andrew 算法求凸包//int ConvexHull(Point *p,int n,Point *ch)
//{
//    int m=0;
//    sort(p,p+n);
//    n=unique(p, p+n)-p;
//
//    for(int i=0;i<n;i++)
//    {
//        while(m>1&&dcmp(Cross(p[m-1]-p[m-2],p[i]-p[m-2]))<=0)  m--;
//        ch[m++]=p[i];
//    }
//
//    int k=m;
//
//    for(int i=n-2;i>=0;i--)
//    {
//       while(m>k&&dcmp(Cross(p[m-1]-p[m-2],p[i]-p[m-2]))<=0)   m--;
//        ch[m++]=p[i];
//    }
//
//    if(n>1) m--;
//
//    return m;
//
//
//}vector<Point> ConvexHull(vector<Point> p) {sort(p.begin(), p.end());p.erase(unique(p.begin(), p.end()), p.end());int n = p.size();int m = 0;vector<Point> ch(n+1);for(int i = 0; i < n; i++) {while(m > 1 && 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-2; i >= 0; i--) {while(m > k && Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]) <= 0) m--;ch[m++] = p[i];}if(n > 1) m--;ch.resize(m);return ch;
}
//Point p[1010];
Point ch[1010];int main()
{int n;double L;cin>>n>>L;vector<Point> p;Point temp;for(int i=0;i<n;i++){temp=read_point();p.push_back(temp);}double ans=PI*L*2;vector<Point> ch=ConvexHull(p);for(int i=0;i<ch.size();i++)ans+=Length(ch[(i+1)%ch.size()]-ch[i]);printf("%.0f\n",ans);}

对于凸包比较喜欢这个写法虽然是一样的

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

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