uva 12296 Pieces and Discs

题意：

有个矩形，左下角(0,0),左上角(L,W）.

思路：

除了圆盘之外，本题的输入也是个PSLG，因此可以按照前面叙述的算法求出各个区域：只需把线段视为直线，用切割凸多边形的方法 ：每次读入线段，切割所有块，最终得到若干凸多边形

如何判断多边形是否与圆盘相交：如果多边形的边和圆周规范相交，圆盘和多边形一定相交，

1：即使完全没有公共点，也可以相交，互相内含 需要判断多边形是否有顶点在园内，还需要判断圆心是否在多边形内；

2.如果不规范：  a:带判断的线段在园外；

b:带判断的线段在园内即两个端点在圆上，只需要判断中点是否在园内；

 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<memory.h>
 #include<cstdlib>
 #include<vector>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define LL long long int
 #define up(i,x,y) for(i=x;i<=y;i++)
 #define w(a) while(a)
 const double inf=0x3f3f3f3f;
 const double eps = 1e-;
 const double PI = acos(-1.0);
 using namespace std;

 double dcmp(double x)
 {
     if(fabs(x) < eps) return ;
     else return x <  ? - : ;
 }

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

 typedef Point Vector;

 typedef vector<Point> Polygon;

 Vector operator + (Vector A, Vector B)
 {
     return Vector(A.x+B.x, A.y+B.y);
 }

 Vector operator - (Point A, Point 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);
 }

 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 - A.y*B.x;
 }
 double Length2(Vector A)
 {
     return Dot(A, A);
 }

 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;
 }

 bool OnSegment(Point p, Point a1, Point a2)
 {
     return dcmp(Cross(a1-p, a2-p)) ==  && dcmp(Dot(a1-p, a2-p)) < ;
 }

 // 多边形的有向面积
 double PolygonArea(Polygon poly)
 {
     double area = ;
     int n = poly.size();
     for(int i = ; i < n-; i++)
         area += Cross(poly[i]-poly[], poly[(i+)%n]-poly[]);
     return area/;
 }

 // cut with directed line A->B, return the left part
 // may return a single point or a line segment
 Polygon CutPolygon(Polygon poly, Point A, Point B)
 {
     Polygon newpoly;
     int n = poly.size();
     for(int i = ; i < n; i++)
     {
         Point C = poly[i];
         Point D = poly[(i+)%n];
         if(dcmp(Cross(B-A, C-A)) >= ) newpoly.push_back(C);
         if(dcmp(Cross(B-A, C-D)) != )
         {
             Point ip = GetLineIntersection(A, B-A, C, D-C);
             if(OnSegment(ip, C, D)) newpoly.push_back(ip);
         }
     }
     return newpoly;
 }

 int isPointInPolygon(Point p, Polygon v)
 {
     int wn = ;
     int n = v.size();
     for(int i = ; i < n; i++)
     {
         if(OnSegment(p, v[i], v[(i+)%n])) return -; // 在边界上
         int k = dcmp(Cross(v[(i+)%n]-v[i], p-v[i]));
         int d1 = dcmp(v[i].y - p.y);
         int d2 = dcmp(v[(i+)%n].y - p.y);
         if(k >  && d1 <=  && d2 > ) wn++;
         if(k <  && d2 <=  && d1 > ) wn--;
     }
     if (wn != ) return ; // 内部
     return ; // 外部
 }

 // 点在圆心内。圆周上不算
 bool isInCircle(Point p, Point center, double R)
 {
     return dcmp(Length2(p-center) - R*R) < ;
 }

 // 直线AB和圆心为C，半径为r的圆的交点
 // 返回交点个数，t1, t2分别为两个交点在直线方程中的参数，p1和p2为交点本身
 int getLineCircleIntersection(Point A, Point B, Point C, double r, double& t1, double& t2)
 {
     // 初始方程：(A.x + t(B.x - A.x) - C.x)^2 + (A.y + t(B.y - A.y) - C.y)^2 = r^2
     // 整理得：(at + b)^2 + (ct + d)^2 = r^2
     double a = B.x - A.x;
     double b = A.x - C.x;
     double c = B.y - A.y;
     double d = A.y - C.y;
     // 展开得：(a^2 + c^2)t^2 + 2(ab + cd)t + b^2 + d^2 - r^2 = 0，即et^2 + ft + g = 0
     double e = a * a + c * c;
     double f =  * (a * b + c * d);
     double g = b * b + d * d - r * r;
     double delta = f * f -  * e * g; // 判别式
     if(dcmp(delta) < ) return ; // 相离
     if(dcmp(delta) == )  // 相切
     {
         t1 = t2 = -f / ( * e);
         return ;
     }
     t1 = (-f - sqrt(delta)) / ( * e);
     t2 = (-f + sqrt(delta)) / ( * e);
     return ;
 }

 // 圆和线段是否相交（相切不算）。线段不考虑端点
 bool CircleIntersectSegment(Point A, Point B, Point p, double R)
 {
     double t1, t2;
     int c = getLineCircleIntersection(A, B, p, R, t1, t2);
     if(c <= ) return false;
     if(dcmp(t1) >  && dcmp(t1-) < ) return true; // 端点在圆上
     if(dcmp(t2) >  && dcmp(t2-) < ) return true;
     return false;
 }

 /////////// 题目相关
 vector<Polygon> pieces, new_pieces;

 void cut(int x1, int y1, int x2, int y2)
 {
     new_pieces.clear();
     for(int i = ; i < pieces.size(); i++)
     {
         Polygon left = CutPolygon(pieces[i], Point(x1, y1), Point(x2, y2));
         Polygon right = CutPolygon(pieces[i], Point(x2, y2), Point(x1, y1));
         if(left.size() >= ) new_pieces.push_back(left);
         if(right.size() >= ) new_pieces.push_back(right);
     }
     pieces = new_pieces;
 }

 bool DiscIntersectPolygon(Polygon poly, Point p, double R)
 {
     if(isPointInPolygon(p, poly) != ) return true;
     if(isInCircle(poly[], p, R)) return true;
     int n = poly.size();
     for(int i = ; i < n; i++)
     {
         if(CircleIntersectSegment(poly[i], poly[(i+)%n], p, R))
         {
             return true; // 不考虑线段端点
         }
         if(isInCircle((poly[i]+poly[(i+)%n])*0.5, p, R))
         {
             return true; // 两个端点都在圆上
         }
     }
     return false;
 }

 void query(Point p, int R)
 {
     vector<double> ans;
     for(int i = ; i < pieces.size(); i++)
     {
         if(DiscIntersectPolygon(pieces[i], p, R))
         {
             ans.push_back(fabs(PolygonArea(pieces[i])));
         }
     }
     printf("%d", ans.size());
     sort(ans.begin(), ans.end());
     for(int i = ; i < ans.size(); i++)
         printf(" %.2lf", ans[i]);
     printf("\n");
 }

 int main()
 {
     int n, m, L, W;
     while(scanf("%d%d%d%d", &n, &m, &L, &W) ==  && n)
     {
         pieces.clear();

         Polygon bbox;
         bbox.push_back(Point(, ));
         bbox.push_back(Point(L, ));
         bbox.push_back(Point(L, W));
         bbox.push_back(Point(, W));
         pieces.push_back(bbox);

         for(int i = ; i < n; i++)
         {
             int x1, y1, x2, y2;
             scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
             cut(x1, y1, x2, y2);
         }

         for(int i = ; i < m; i++)
         {
             int x, y, R;
             scanf("%d%d%d", &x, &y, &R);
             query(Point(x, y), R);
         }
         printf("\n");
     }
     return ;
 }