再来一道测半平面交模板题 Poj1279 Art Gallery

地址：http://poj.org/problem?id=1279

题目：

　　　　　　　　　　　　　　　　　　　　　　　　Art Gallery

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7329		Accepted: 2938

Description

The art galleries of the new and very futuristic building of the Center for Balkan Cooperation have the form of polygons (not necessarily convex). When a big exhibition is organized, watching over all of the pictures is a big security concern. Your task is that for a given gallery to write a program which finds the surface of the area of the floor, from which each point on the walls of the gallery is visible. On the figure 1. a map of a gallery is given in some co-ordinate system. The area wanted is shaded on the figure 2. 

Input

The number of tasks T that your program have to solve will be on the first row of the input file. Input data for each task start with an integer N, 5 <= N <= 1500. Each of the next N rows of the input will contain the co-ordinates of a vertex of the polygon ? two integers that fit in 16-bit integer type, separated by a single space. Following the row with the co-ordinates of the last vertex for the task comes the line with the number of vertices for the next test and so on.

Output

For each test you must write on one line the required surface - a number with exactly two digits after the decimal point (the number should be rounded to the second digit after the decimal point).

Sample Input

1
7
0 0
4 4
4 7
9 7
13 -1
8 -6
4 -4

Sample Output

80.00
思路：没什么好说的，和前面几题一样都是用来测模板的题，不过还是wa了两次，因为把题目看成是要四舍五入到第二位小数（盲人acmer）

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <algorithm>

 using namespace std;
 const double eps = 1e-;
 //点
 class Point
 {
 public:
     double x, y;

     Point(){}
     Point(double x, double y):x(x),y(y){}

     bool operator < (const Point &_se) const
     {
         return x<_se.x || (x==_se.x && y<_se.y);
     }
     /*******判断ta与tb的大小关系*******/
     static int sgn(double ta,double tb)
     {
         if(fabs(ta-tb)<eps)return ;
         if(ta<tb)   return -;
         return ;
     }
     static double xmult(const Point &po, const Point &ps, const Point &pe)
     {
         return (ps.x - po.x) * (pe.y - po.y) - (pe.x - po.x) * (ps.y - po.y);
     }
     friend Point operator + (const Point &_st,const Point &_se)
     {
         return Point(_st.x + _se.x, _st.y + _se.y);
     }
     friend Point operator - (const Point &_st,const Point &_se)
     {
         return Point(_st.x - _se.x, _st.y - _se.y);
     }
     //点位置相同(double类型)
     bool operator == (const Point &_off) const
     {
         return  Point::sgn(x, _off.x) ==  && Point::sgn(y, _off.y) == ;
     }
     //点位置不同(double类型)
     bool operator != (const Point &_Off) const
     {
         return ((*this) == _Off) == false;
     }
     //两点间距离的平方
     static double dis2(const Point &_st,const Point &_se)
     {
         return (_st.x - _se.x) * (_st.x - _se.x) + (_st.y - _se.y) * (_st.y - _se.y);
     }
     //两点间距离
     static double dis(const Point &_st, const Point &_se)
     {
         return sqrt((_st.x - _se.x) * (_st.x - _se.x) + (_st.y - _se.y) * (_st.y - _se.y));
     }
 };
 //两点表示的向量
 class Line
 {
 public:
     Point s, e;//两点表示，起点[s]，终点[e]
     double a, b, c;//一般式,ax+by+c=0

     Line(){}
     Line(const Point &s, const Point &e):s(s),e(e){}
     Line(double _a,double _b,double _c):a(_a),b(_b),c(_c){}

     //向量与点的叉乘,参数：点[_Off]
     //[点相对向量位置判断]
     double operator /(const Point &_Off) const
     {
         return (_Off.y - s.y) * (e.x - s.x) - (_Off.x - s.x) * (e.y - s.y);
     }
     //向量与向量的叉乘,参数：向量[_Off]
     friend double operator /(const Line &_st,const Line &_se)
     {
         return (_st.e.x - _st.s.x) * (_se.e.y - _se.s.y) - (_st.e.y - _st.s.y) * (_se.e.x - _se.s.x);
     }
     friend double operator *(const Line &_st,const Line &_se)
     {
         return (_st.e.x - _st.s.x) * (_se.e.x - _se.s.x) - (_st.e.y - _st.s.y) * (_se.e.y - _se.s.y);
     }
     //从两点表示转换为一般表示
     //a=y2-y1,b=x1-x2,c=x2*y1-x1*y2
     bool pton()
     {
         a = e.y - s.y;
         b = s.x - e.x;
         c = e.x * s.y - e.y * s.x;
         return true;
     }

     //-----------点和直线（向量）-----------
     //点在向量左边（右边的小于号改成大于号即可,在对应直线上则加上=号）
     //参数：点[_Off],向量[_Ori]
     friend bool operator<(const Point &_Off, const Line &_Ori)
     {
         return (_Ori.e.y - _Ori.s.y) * (_Off.x - _Ori.s.x)
             < (_Off.y - _Ori.s.y) * (_Ori.e.x - _Ori.s.x);
     }

     //点在直线上,参数：点[_Off]
     bool lhas(const Point &_Off) const
     {
         return Point::sgn((*this) / _Off, ) == ;
     }
     //点在线段上,参数：点[_Off]
     bool shas(const Point &_Off) const
     {
         return lhas(_Off)
             && Point::sgn(_Off.x - min(s.x, e.x), ) >  && Point::sgn(_Off.x - max(s.x, e.x), ) <
             && Point::sgn(_Off.y - min(s.y, e.y), ) >  && Point::sgn(_Off.y - max(s.y, e.y), ) < ;
     }

     //点到直线/线段的距离
     //参数： 点[_Off], 是否是线段[isSegment](默认为直线)
     double dis(const Point &_Off, bool isSegment = false)
     {
         ///化为一般式
         pton();

         //到直线垂足的距离
         double td = (a * _Off.x + b * _Off.y + c) / sqrt(a * a + b * b);

         //如果是线段判断垂足
         if(isSegment)
         {
             double xp = (b * b * _Off.x - a * b * _Off.y - a * c) / ( a * a + b * b);
             double yp = (-a * b * _Off.x + a * a * _Off.y - b * c) / (a * a + b * b);
             double xb = max(s.x, e.x);
             double yb = max(s.y, e.y);
             double xs = s.x + e.x - xb;
             double ys = s.y + e.y - yb;
             if(xp > xb + eps || xp < xs - eps || yp > yb + eps || yp < ys - eps)
                 td = min(Point::dis(_Off,s), Point::dis(_Off,e));
         }

         return fabs(td);
     }

     //关于直线对称的点
     Point mirror(const Point &_Off) const
     {
         ///注意先转为一般式
         Point ret;
         double d = a * a + b * b;
         ret.x = (b * b * _Off.x - a * a * _Off.x -  * a * b * _Off.y -  * a * c) / d;
         ret.y = (a * a * _Off.y - b * b * _Off.y -  * a * b * _Off.x -  * b * c) / d;
         return ret;
     }
     //计算两点的中垂线
     static Line ppline(const Point &_a, const Point &_b)
     {
         Line ret;
         ret.s.x = (_a.x + _b.x) / ;
         ret.s.y = (_a.y + _b.y) / ;
         //一般式
         ret.a = _b.x - _a.x;
         ret.b = _b.y - _a.y;
         ret.c = (_a.y - _b.y) * ret.s.y + (_a.x - _b.x) * ret.s.x;
         //两点式
         if(std::fabs(ret.a) > eps)
         {
             ret.e.y = 0.0;
             ret.e.x = - ret.c / ret.a;
             if(ret.e == ret. s)
             {
                 ret.e.y = 1e10;
                 ret.e.x = - (ret.c - ret.b * ret.e.y) / ret.a;
             }
         }
         else
         {
             ret.e.x = 0.0;
             ret.e.y = - ret.c / ret.b;
             if(ret.e == ret. s)
             {
                 ret.e.x = 1e10;
                 ret.e.y = - (ret.c - ret.a * ret.e.x) / ret.b;
             }
         }
         return ret;
     }

     //------------直线和直线（向量）-------------
     //直线重合,参数：直线向量[_st],[_se]
     static bool equal(const Line &_st, const Line &_se)
     {
         return _st.lhas(_se.e) && _se.lhas(_se.s);
     }
     //直线平行，参数：直线向量[_st],[_se]
     static bool parallel(const Line &_st,const Line &_se)
     {
         return Point::sgn(_st / _se, ) == ;
     }
     //两直线（线段）交点，参数：直线向量[_st],[_se],交点
     //返回-1代表平行，0代表重合，1代表相交
     static bool crossLPt(const Line &_st,const Line &_se,Point &ret)
     {
         if(Line::parallel(_st,_se))
         {
             if(Line::equal(_st,_se)) return ;
             return -;
         }
         ret = _st.s;
         double t = (Line(_st.s,_se.s)/_se)/(_st/_se);
         ret.x += (_st.e.x - _st.s.x) * t;
         ret.y += (_st.e.y - _st.s.y) * t;
         return ;
     }
     //------------线段和直线（向量）----------
     //线段和直线交
     //参数：直线[_st],线段[_se]
     friend bool crossSL(const Line &_st,const Line &_se)
     {
         return Point::sgn((_st / _se.s) * (_st / _se.e) ,) <= ;
     }

     //------------线段和线段（向量）----------
     //判断线段是否相交(注意添加eps)，参数：线段[_st],线段[_se]
     static bool isCrossSS(const Line &_st,const Line &_se)
     {
         //1.快速排斥试验判断以两条线段为对角线的两个矩形是否相交
         //2.跨立试验（等于0时端点重合）
         return
             max(_st.s.x, _st.e.x) >= min(_se.s.x, _se.e.x) &&
             max(_se.s.x, _se.e.x) >= min(_st.s.x, _st.e.x) &&
             max(_st.s.y, _st.e.y) >= min(_se.s.y, _se.e.y) &&
             max(_se.s.y, _se.e.y) >= min(_st.s.y, _st.e.y) &&
             Point::sgn((_st / Line(_st.s, _se.s)) * (_st / Line(_st.s, _se.e)), ) <=  &&
             Point::sgn((_se / Line(_se.s, _st.s)) * (_se / Line(_se.s, _st.e)), ) <= ;
     }
 };
 class Polygon
 {
 public:
     const static int maxpn = ;
     Point pt[maxpn];//点（顺时针或逆时针）
     int n;//点的个数

     Point& operator[](int _p)
     {
         return pt[_p];
     }

     //求多边形面积，多边形内点必须顺时针或逆时针
     double area() const
     {
         double ans = 0.0;
         for(int i = ; i < n; i ++)
         {
             int nt = (i + ) % n;
             ans += pt[i].x * pt[nt].y - pt[nt].x * pt[i].y;
         }
         return fabs(ans / 2.0);
     }
     //求多边形重心，多边形内点必须顺时针或逆时针
     Point gravity() const
     {
         Point ans;
         ans.x = ans.y = 0.0;
         double area = 0.0;
         for(int i = ; i < n; i ++)
         {
             int nt = (i + ) % n;
             double tp = pt[i].x * pt[nt].y - pt[nt].x * pt[i].y;
             area += tp;
             ans.x += tp * (pt[i].x + pt[nt].x);
             ans.y += tp * (pt[i].y + pt[nt].y);
         }
         ans.x /=  * area;
         ans.y /=  * area;
         return ans;
     }
     //判断点在凸多边形内，参数：点[_Off]
     bool chas(const Point &_Off) const
     {
         double tp = , np;
         for(int i = ; i < n; i ++)
         {
             np = Line(pt[i], pt[(i + ) % n]) / _Off;
             if(tp * np < -eps)
                 return false;
             tp = (fabs(np) > eps)?np: tp;
         }
         return true;
     }
     //判断点是否在任意多边形内[射线法]，O(n)
     bool ahas(const Point &_Off) const
     {
         int ret = ;
         double infv = 1e-;//坐标系最大范围
         Line l = Line(_Off, Point( -infv ,_Off.y));
         for(int i = ; i < n; i ++)
         {
             Line ln = Line(pt[i], pt[(i + ) % n]);
             if(fabs(ln.s.y - ln.e.y) > eps)
             {
                 Point tp = (ln.s.y > ln.e.y)? ln.s: ln.e;
                 if(fabs(tp.y - _Off.y) < eps && tp.x < _Off.x + eps)
                     ret ++;
             }
             else if(Line::isCrossSS(ln,l))
                 ret ++;
         }
         return (ret %  == );
     }
     //凸多边形被直线分割,参数：直线[_Off]
     Polygon split(Line _Off)
     {
         //注意确保多边形能被分割
         Polygon ret;
         Point spt[];
         double tp = 0.0, np;
         bool flag = true;
         int i, pn = , spn = ;
         for(i = ; i < n; i ++)
         {
             if(flag)
                 pt[pn ++] = pt[i];
             else
                 ret.pt[ret.n ++] = pt[i];
             np = _Off / pt[(i + ) % n];
             if(tp * np < -eps)
             {
                 flag = !flag;
                 Line::crossLPt(_Off,Line(pt[i], pt[(i + ) % n]),spt[spn++]);
             }
             tp = (fabs(np) > eps)?np: tp;
         }
         ret.pt[ret.n ++] = spt[];
         ret.pt[ret.n ++] = spt[];
         n = pn;
         return ret;
     }

     /** 卷包裹法求点集凸包，_p为输入点集，_n为点的数量 **/
     void ConvexClosure(Point _p[],int _n)
     {
         sort(_p,_p+_n);
         n=;
         for(int i=;i<_n;i++)
         {
             while(n>&&Point::sgn(Line(pt[n-],pt[n-])/Line(pt[n-],_p[i]),)<=)
                 n--;
             pt[n++]=_p[i];
         }
         int _key=n;
         for(int i=_n-;i>=;i--)
         {
             while(n>_key&&Point::sgn(Line(pt[n-],pt[n-])/Line(pt[n-],_p[i]),)<=)
                 n--;
             pt[n++]=_p[i];
         }
         if(n>)   n--;//除去重复的点，该点已是凸包凸包起点
     }
 //    /****** 寻找凸包的graham 扫描法********************/
 //    /****** _p为输入的点集,_n为点的数量****************/
 //    /**使用时需把gmp函数放在类外，并且看情况修改pt[0]**/
 //    bool gcmp(const Point &ta,const Point &tb)/// 选取与最后一条确定边夹角最小的点，即余弦值最大者
 //    {
 //        double tmp=Line(pt[0],ta)/Line(pt[0],tb);
 //        if(Point::sgn(tmp,0)==0)
 //            return Point::dis(pt[0],ta)<Point::dis(pt[0],tb);
 //        else if(tmp>0)
 //            return 1;
 //        return 0;
 //    }
 //    void graham(Point _p[],int _n)
 //    {
 //        int cur=0;
 //        for(int i=1;i<_n;i++)
 //            if(Point::sgn(_p[cur].y,_p[i].y)>0 || (Point::sgn(_p[cur].y,_p[i].y)==0 && Point::sgn(_p[cur].x,_p[i].x)>0))
 //                cur=i;
 //        swap(_p[cur],_p[0]);
 //        n=0,pt[n++]=_p[0];
 //        if(_n==1)   return;
 //        sort(_p+1,_p+_n,Polygon::gcmp);
 //        pt[n++]=_p[1],pt[n++]=_p[2];
 //        for(int i=3;i<_n;i++)
 //        {
 //            while(Point::sgn(Line(pt[n-2],pt[n-1])/Line(pt[n-2],_p[i]),0)<0)
 //                n--;
 //            pt[n++]=_p[i];
 //        }
 //    }
     //凸包旋转卡壳(注意点必须顺时针或逆时针排列)
     //返回值凸包直径的平方（最远两点距离的平方）
     double rotating_calipers()
     {
         int i = ;
         double ret = 0.0;
         pt[n] = pt[];
         for(int j = ; j < n; j ++)
         {
             while(fabs(Point::xmult(pt[i+],pt[j], pt[j + ])) > fabs(Point::xmult(pt[i],pt[j], pt[j + ])) + eps)
                 i = (i + ) % n;
             //pt[i]和pt[j],pt[i + 1]和pt[j + 1]可能是对踵点
             ret = (ret, max(Point::dis(pt[i],pt[j]), Point::dis(pt[i + ],pt[j + ])));
         }
         return ret;
     }

     //凸包旋转卡壳(注意点必须逆时针排列)
     //返回值两凸包的最短距离
     double rotating_calipers(Polygon &_Off)
     {
         int i = ;
         double ret = 1e10;//inf
         pt[n] = pt[];
         _Off.pt[_Off.n] = _Off.pt[];
         //注意凸包必须逆时针排列且pt[0]是左下角点的位置
         while(_Off.pt[i + ].y > _Off.pt[i].y)
             i = (i + ) % _Off.n;
         for(int j = ; j < n; j ++)
         {
             double tp;
             //逆时针时为 >,顺时针则相反
             while((tp = Point::xmult(_Off.pt[i + ],pt[j], pt[j + ]) - Point::xmult(_Off.pt[i], pt[j], pt[j + ])) > eps)
                 i = (i + ) % _Off.n;
             //(pt[i],pt[i+1])和(_Off.pt[j],_Off.pt[j + 1])可能是最近线段
             ret = min(ret, Line(pt[j], pt[j + ]).dis(_Off.pt[i], true));
             ret = min(ret, Line(_Off.pt[i], _Off.pt[i + ]).dis(pt[j + ], true));
             if(tp > -eps)//如果不考虑TLE问题最好不要加这个判断
             {
                 ret = min(ret, Line(pt[j], pt[j + ]).dis(_Off.pt[i + ], true));
                 ret = min(ret, Line(_Off.pt[i], _Off.pt[i + ]).dis(pt[j], true));
             }
         }
         return ret;
     }

     //-----------半平面交-------------
     //复杂度:O(nlog2(n))
     //#include <algorithm>
     //半平面计算极角函数[如果考虑效率可以用成员变量记录]
     static double hpc_pa(const Line &_Off)
     {
         return atan2(_Off.e.y - _Off.s.y, _Off.e.x - _Off.s.x);
     }
     //半平面交排序函数[优先顺序: 1.极角 2.前面的直线在后面的左边]
     static bool hpc_cmp(const Line &l, const Line &r)
     {
         double lp = hpc_pa(l), rp = hpc_pa(r);
         if(fabs(lp - rp) > eps)
             return lp < rp;
         return Point::xmult(r.s,l.s, r.e) < -eps;
     }
     static int judege(const Line &_lx,const Line &_ly,const Line &_lz)
     {
         Point tmp;
         Line::crossLPt(_lx,_ly,tmp);
         return Point::sgn(Point::xmult(_lz.s,tmp,_lz.e),);
     }
     //获取半平面交的多边形（多边形的核）
     //参数：向量集合[l]，向量数量[ln];(半平面方向在向量左边）
     //函数运行后如果n[即返回多边形的点数量]为0则不存在半平面交的多边形（不存在区域或区域面积无穷大）
     Polygon& halfPanelCross(Line _Off[], int ln)
     {
         Line dequeue[maxpn];//用于计算的双端队列
         int i, tn;
         sort(_Off, _Off + ln, hpc_cmp);
         //平面在向量左边的筛选
         for(i = tn = ; i < ln; i ++)
             if(fabs(hpc_pa(_Off[i]) - hpc_pa(_Off[i - ])) > eps)
                 _Off[tn ++] = _Off[i];
         ln = tn,n = ;
         int bot = , top = ;
         dequeue[] = _Off[];
         dequeue[] = _Off[];
         for(i = ; i < ln; i ++)
         {
             while(bot < top &&  Polygon::judege(dequeue[top],dequeue[top-],_Off[i]) > )
                 top --;
             while(bot < top &&  Polygon::judege(dequeue[bot],dequeue[bot+],_Off[i]) > )
                 bot ++;
             dequeue[++ top] = _Off[i];
         }

         while(bot < top && Polygon::judege(dequeue[top],dequeue[top-],dequeue[bot]) > )
             top --;
         while(bot < top && Polygon::judege(dequeue[bot],dequeue[bot+],dequeue[top]) > )
             bot ++;
         //计算交点(注意不同直线形成的交点可能重合)
         if(top <= bot + )
             return (*this);
         for(i = bot; i < top; i ++)
             Line::crossLPt(dequeue[i],dequeue[i + ],pt[n++]);
         if(bot < top + )
             Line::crossLPt(dequeue[bot],dequeue[top],pt[n++]);
         return (*this);
     }
 };

 int n,t;
 Point pt[];
 Line ln[];
 Polygon ans;
 int main(void)
 {
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(int i=;i<n;i++)
             scanf("%lf%lf",&pt[i].x,&pt[i].y);
         pt[n++]=pt[];
         for(int i=n-;i;i--)
             ln[i-]=Line(pt[i],pt[i-]);
         //for(int i=0;i<n-1;i++)
         //    printf("%.2f %.2f %.2f %.2f\n",ln[i].s.x,ln[i].s.y,ln[i].e.x,ln[i].e.y);
         ans.halfPanelCross(ln,n-);
         double area=;
         if(ans.n==)
             area=;
         else
             area=ans.area();
         printf("%.2f\n",area);
     }
     return ;
 }