PSLG,直线切割凸多边形，和判断圆与多边形相交

分析：

除了圆盘之外，本题的输入也是一个PSLG,因此可以按照前面叙述的算法求出各个区域。但由于本题的特殊性，不难发现把线段改成直线后答案不变，因此每个块都是凸多边形，

可以用切割凸多边形的方法求解：每读入一条线段，都把它当做直线，切割所有块。这样，我们最终得到了若干凸多边形，需要分别判断是否与圆盘相交。

如何让判断多边形是否和圆盘相交？，显然，如果多边形的边和圆周规范相交，圆盘和多变性一定相交，但反过来却不成立——圆盘和多边形相交，多边形的边和圆周不一定规范相交。

（1）即使完全没有公共点的时候，圆盘和多边形也可以相交，原因是二者可以相互内含。因此，需要判断多边形是否有顶点在圆内，还需要判断圆心是否在多边形内。

（2）如果是非规范相交,需要分情况讨论。在图中，待判断的线段（用粗线表示）完全在圆外；在图中待判断的线段则是完全在内部。判断方法很简单，只需判断线段中点是否在圆内即可。

直接切割多边形~    判断多边形和园盘是否有公共点（面积>0） 

1  内含的情况--只要多边形poly[0] 在圆内、或者圆心在多边形内

2  相交的情况-如果不是规范相交，那么不是内含，却有非零公共面积只有一种情况，就是两个点都在圆上，只有判断中点在圆上即可。

 每一个案例忘记输出空行  并不提示Presentation Error ，wa每次pieces更新的时候，newpieces 需要清零

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<cmath>#include<cctype>#include<string>#include<set>#include<map>#include<queue>#include<stack>#include<vector>using namespace std;const double eps=1e-6;int dcmp(double x){    if(fabs(x)<eps) return 0;    return x>0?1:-1;    //return fabs(x) < eps ? 0 : (x > 0 ? 1 : -1);}struct point{  double x;  double y;  point(){}  point(double x,double  y):x(x),y(y){}  void in()  {      cin>>x>>y;  }  void out()  {      cout<<x<<' '<<y<<endl;  }  point operator + (const point &t) const  {      return point(x+t.x,y+t.y);  }  point operator - (const point &t) const  {      return point(x-t.x,y-t.y);  }  point operator * (const double &t) const  {      return point(x*t,y*t);  }  point operator / (const double &t) const  {      return point(x/t,y/t);  }  bool operator < (const point &t) const  {      return (dcmp(x-t.x)<0||(dcmp(x-t.x)==0&&dcmp(y-t.y)<0));  }  bool operator == (const point &t) const  {      return dcmp(x-t.x) ==0 &&dcmp(y-t.y)==0;  }};double cross(point a,point b){    return a.x*b.y-a.y*b.x;}double dot(point a,point b){    return a.x*b.x+a.y*b.y;}double length(point a){    return sqrt(dot(a,a));}point nomal(point t){    double l=length(t);    return  point(-t.y/l,t.x/l);}struct line{    point p;    point v;    double ang;    line() {}    line(point p,point v):p(p),v(v){        ang=atan2(v.y,v.x);    }    bool operator < (const line &l) const    {        return ang<l.ang;    }    point ppoint(double t)    {        return point(p+v*t);    }};struct Circle{    point c;    double r;    Circle(point c=point(0,0),double r=0):c(c),r(r) {}    point ppoint(double a)    {        return point(c.x+r*cos(a),c.y+r*sin(a));    }};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.ppoint(t1));        return 1;    }    else    {        t1=(-f-sqrt(delta))/(2*e);        t2=(-f+sqrt(delta))/(2*e);        sol.push_back(l.ppoint(t1));        sol.push_back(l.ppoint(t2));        return 2;    }}bool onleft(line l,point p){    return cross(l.v,p-l.p)>0;}point getintersection(line a,line b){    point u=a.p-b.p;    double t=cross(b.v,u)/cross(a.v,b.v);    return a.p+a.v*t;}bool sgementproperintersection(point a1,point a2,point b1,point b2){    double c1=cross(a2-a1,b1-a1),c2=cross(a2-a1,b2-a1),           c3=cross(b2-b1,a1-b1),c4=cross(b2-b1,a2-b1);    return dcmp(c1)*dcmp(c2)<0&&dcmp(c3)*dcmp(c4)<0;}bool onsegment(point p,point a1,point a2){    return dcmp(cross(a1-p,a2-p))==0&&dcmp(dot(a1-p,a2-p))<0;}typedef vector<point> Polygon;double PolygonArea(Polygon poly){    double area=0;    int n=poly.size();    for(int i=1;i<n-1;i++)    {        area+=cross(poly[i]-poly[0],poly[(i+1)%n]-poly[0]);    }    return area/2;}vector<Polygon> pieces,newpieces;Polygon CutPolygon(Polygon poly,point a,point b){    Polygon newPoly;    int n=poly.size();    for(int i=0;i<n;i++)    {        point c=poly[i];        point d=poly[(i+1)%n];        if(dcmp(cross(b-a,c-a))>=0) newPoly.push_back(c);        if(dcmp(cross(b-a,d-c))!=0)        {            point ip=getintersection(line(c,d-c),line(a,b-a));            if(onsegment(ip,c,d)) newPoly.push_back(ip); //此时必须用不含端点的“在线段上"        }    }    return newPoly;}void cut(point a,point b){    newpieces.clear();  //仅仅是一个temp 记得清空    for(int i=0;i<pieces.size();i++)    {        Polygon poly=pieces[i];        Polygon left=CutPolygon(poly,a,b);        Polygon right=CutPolygon(poly,b,a);        if(left.size()>=3) newpieces.push_back(left);        if(right.size()>=3) newpieces.push_back(right);    }    pieces=newpieces;}bool isPointInPolygon(point p,Polygon poly){    int wn=0;    int n=poly.size();    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;}double length2(point a){    return dot(a,a);}bool inCircle(Circle  C,point p) //圆周不算{    if(dcmp(length2(p-C.c)-C.r*C.r)<0) return 1;    else return 0;}bool CircleSegIntersection(Circle C,point a,point b) //线段端点不算{    double t1,t2;    vector<point> sol;    if(getLineCircleIntersection(line(a,b-a),C,t1,t2,sol)<=1) return 0;    if(dcmp(t1)>0&&dcmp(t1-1)<0) return 1;    if(dcmp(t2)>0&&dcmp(t2-1)<0) return 1;    return 0;}bool CirclePolyIntersection(Circle C,Polygon poly){    if(isPointInPolygon(C.c,poly)) return 1;    if(inCircle(C,poly[0]))  return 1;    point a,b;    int n=poly.size();    for(int i=0;i<n;i++)    {        a=poly[i];        b=poly[(i+1)%n];        if(CircleSegIntersection(C,a,b)) return 1;        if(inCircle(C,(a+b)/2)) return 1;    }    return 0;}void Query(Circle circle){    vector<double> ans;    for(int i=0;i<pieces.size();i++)        if(CirclePolyIntersection(circle,pieces[i]))    {        ans.push_back(fabs(PolygonArea(pieces[i])));    }    sort(ans.begin(),ans.end());    cout<<ans.size();    for(int i=0;i<ans.size();i++)    {        printf(" %.2f",ans[i]);    }    cout<<endl;}int main(){    int n,m,l,w;    Circle C;    while(cin>>n>>m>>l>>w)    {        if(!n) break;        pieces.clear();        Polygon bbox;        bbox.push_back(point(0,0));        bbox.push_back(point(l,0));        bbox.push_back(point(l,w));        bbox.push_back(point(0,w));        pieces.push_back(bbox);        point a,b;        for(int i=0;i<n;i++)        {            a.in();            b.in();            cut(a,b);        }        for(int i=0;i<m;i++)        {            cin>>C.c.x>>C.c.y>>C.r;            Query(C);        }        printf("\n");    }    return 0;}