2015 北京网络赛 E Border Length hihoCoder 1231 树状数组 (2015-11-05 09:30)

#include <iostream>
#include <algorithm>
#include <string.h>
#include <math.h>
#include <cstdio>
using namespace std;
const double eps=0.000000000001;
const double PI=acos(-1.0);
const int maxn=3005;
int dcmp(double a)
{
    if(fabs(a)<eps)return 0;
    else return a>0?1:-1;
}
struct Point
{
  double x,y;
  Point(double cx=0,double cy=0)
  {
      x=cx; y=cy;
  }
  double ang;
  void Angle()
  {
      ang=atan2(y,x);
      if(ang<0){
        ang+=2*PI;
      }
  }
  bool operator <(const Point &rhs)const
  {
     return ang<rhs.ang;
  }
};
struct Circle
{
    Point c;
    double r;
    Circle(Point cc=Point(0,0),double cr=0)
    {
        c=cc; r=cr;
    }
    Point point(double a)
    {
        return Point(c.x+cos(a)*r, c.y+sin(a)*r);
    }
};
Point operator -(Point A, Point B)
{
    return Point(A.x-B.x,A.y-B.y);
}
Point operator +(Point A, Point B)
{
    return Point(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(Point A,Point B)
{
    return A.x*B.x+A.y*B.y;
}
double Length(Point A)
{
     return sqrt(Dot(A,A));
}
double Cross(Point A, Point B)
{
    return A.x*B.y - A.y*B.x;
}
Point TE[maxn];
Circle C;
struct Ins{
  Point c;
  int id;
}Poi[maxn];
struct Line
{
    Point p;
    Point v;
    Line(Point cp=Point(0,0),Point cv=Point(0,0))
    {
        p=cp; v=cv;
    }
    Point point(double a)
    {
        Point te=Point(v.x*a,v.y*a);
        return p+te;
    }
};
int getLineCicleIntersection(Line L,Circle C, double &t1,double &t2,Point &A,Point &B)
{
    double a=L.v.x,
           b=L.p.x-C.c.x,
           c=L.v.y,
           d=L.p.y-C.c.y;
    double e=a*a+c*c,
           f=2*(a*b+c*d),
           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);A=L.point(t1);
        return 1;
    }
    t1=(-f - sqrt(delta))/(2*e);
    A=L.point(t1);
    t2=(-f + sqrt(delta))/(2*e);
    B=L.point(t2);
    return 2;
}
int judjiaodian(int n)
{
     int nu=0;
     double t1,t2;
     Point A,B;
     for(int i=0; i<n; i++)
     {
         int ge=getLineCicleIntersection(Line(TE[i],TE[i+1]-TE[i]),C,t1,t2,A,B);//计算直线和圆交点 大白书模板
           if(ge==0)continue;
           double s=Length(TE[i]-TE[i+1]);
           double d1=Length(TE[i]-A);
           double d2=Length(TE[i+1]-A);
          if(ge==1){
             if(dcmp(s-d1)>=0&&dcmp(s-d2)>=0) //判断是否在这条边上
             {
                   Poi[nu].c=A; Poi[nu].id=i; nu++;
             }
            continue;
          }
          double d3=Length(TE[i]-B);
          double d4=Length(TE[i+1]-B);
          if(dcmp(s-d1)>=0&&dcmp(s-d2)>=0&&dcmp(s-d3)>=0&&dcmp(s-d4)>=0)//如果存在两个点
            {
                 if(d1>d3){//应该把离起点进的点放在前面
                    swap(A,B);
                 }
                 Poi[nu].c=A; Poi[nu].id=i;nu++;
                 Poi[nu].c=B; Poi[nu].id=i;nu++;
            }else if(dcmp(s-d1)>=0&&dcmp(s-d2)>=0)
            {
                 Poi[nu].c=A; Poi[nu].id=i;nu++;
            }else if(dcmp(s-d3)>=0&&dcmp(s-d4)>=0)
            {
                 Poi[nu].c=B; Poi[nu].id=i;nu++;
            }
     }
     int ge=1;
     for(int i=1; i<nu; i++)
        if( (Poi[i].c==Poi[ge-1].c) == false )
         Poi[ge++]=Poi[i];
     nu=ge;
     return nu;
}
int isPointInpolygon(Point p,int n)//判断点是否在多边形内
{
    int wn=0;
    for(int i=0; i<n; i++)
    {
        int k=dcmp(Cross(TE[i+1]-TE[i],p-TE[i]));
        int d1=dcmp(TE[i].y -p.y );
        int d2=dcmp(TE[i+1].y-p.y);
        if( k > 0 && d1 <= 0 && d2 > 0 )wn++;
        if( k < 0 && d2 <= 0 && d1 >0  )wn--;
     }
     if(wn!=0)return 1;
     return 0;
}
void solve(int n)
{
    int numOfInCir=0;
    double dist=0;
    for(int i=0; i<n; i++)
        {
            if( dcmp( C.r-Length(TE[i]-C.c) )>=0 )numOfInCir++;
            dist+=Length( TE[i+1]-TE[i] );
        }
    if(isPointInpolygon(C.c,n))
    {
        if(numOfInCir==n){
            printf("%.0lf\n",dist);
        }else printf("%.0lf\n",PI*2*C.r);
        return;
    }
    if(numOfInCir == n)
    {
            printf("%.0lf\n",dist);
    }else printf("0\n");
}
Point Hu[maxn];
double judsolveCir(int nu,int n)
{
    for(int i=0; i<nu; i++)
    {
        Hu[i]=Poi[i].c-C.c;
        Hu[i].Angle();
    }
    sort(Hu,Hu+nu);
    Hu[nu]=Hu[0];
    Hu[nu].ang+=2*PI;
    double ans=0;
    for(int i=0; i<nu; i++)
    {
        double ang=(Hu[i].ang+Hu[i+1].ang)/2;
        if(ang>2*PI)ang-=2*PI;
        Point t =C.point(ang);
        if(isPointInpolygon(t,n))
        {
           ang=Hu[i+1].ang-Hu[i].ang;
           ans+=ang*C.r;
        }
    }
    return ans;
}
Point TP[maxn];
double judsolvePoly(int nu, int n)
{
   int loc=0,num=0;
   for(int i=0; i<n; i++)//融合那些点
   {
       TP[num++]=TE[i];
       while(loc<nu&&Poi[loc].id<=i){
         TP[num++]=Poi[loc].c; loc++;
       }
   }
   loc=1;
   for(int i=1; i<num; i++)
    if((TP[i]==TP[loc-1])==false )TP[loc++]=TP[i];
   num=loc;
   TP[num]=TP[0];
   double ans=0;
   for(int i=0; i<num; i++)
   {
       Point tt=TP[i]+TP[i+1];
       tt.x*=0.5;
       tt.y*=0.5;
       double d1=Length(C.c-tt);
       if(dcmp(C.r-d1)>=0){
         ans+=Length(TP[i+1]-TP[i]);
       }
   }
   return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)==1&&n)
    {
        for(int i=0; i<n; i++)
            scanf("%lf%lf",&TE[i].x,&TE[i].y);
            TE[n]=TE[0];
        scanf("%lf%lf%lf",&C.c.x,&C.c.y,&C.r);
        int nu=judjiaodian(n);//计算交点
        if(nu==0||nu==1){
           solve(n);// 特判那几种情况
           continue;
        }
        double ans=judsolveCir(nu,n);// 计算圆在多边形内的部分
        ans+=judsolvePoly(nu,n);// 计算多边形在圆内的部分
        printf("%.0lf\n",ans);
    }
    return 0;
}