uva 1308 - Viva Confetti

这个题目的方法是将圆盘分成一个个圆环，然后判断这些圆环是否被上面的圆覆盖；

如果这个圆的圆周上的圆弧都被上面的覆盖，暂时把它标记为不可见；

然后如果他的头上有个圆，他有个圆弧可见，那么他自己本身可见，并且可以把这条圆弧下面的第一个圆重新标记为可见；

另外，圆弧可见还是不可见利用它的中点来进行判断；

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <algorithm>
 #include <vector>

 using namespace std;

 const int maxn =  + ;
 const double eps = 1e-;        //别开太大，样例数据就到达1e-11级别
 const double pi = acos(-);

 int dcmp(double x)
 {
     return fabs(x) < eps ?  : (x >  ?  : -);
 }

 struct Point
 {
     double x;
     double y;

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

     bool operator < (const Point& e) const
     {
         return dcmp(x - e.x) <  || (dcmp(x - e.x) ==  && dcmp(y - e.y) < );
     }

     bool operator == (const Point& e) const
     {
         return dcmp(x - e.x) ==  && dcmp(y - e.y) == ;
     }

     int read()
     {
         return scanf("%lf%lf", &x, &y);
     }
 };

 typedef Point Vector;

 Vector operator + (Point A, Point 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 * (Point A, double p)
 {
     return Vector(A.x * p, A.y * p);
 }

 Vector operator / (Point A, double p)
 {
     return Vector(A.x / p, A.y / p);
 }

 struct Circle
 {
     Point c;
     double r;

     Circle() {}
     Circle(Point c, double r):c(c), r(r) {}

     int read()
     {
         return scanf("%lf%lf%lf", &c.x, &c.y, &r);
     }

     Point point(double a)
     {
         return Point(c.x + r * cos(a), c.y + r * sin(a));
     }
 };

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

 double Length(Vector A)
 {
     return sqrt(Dot(A, A));
 }

 double angle(Vector v)      //求向量的极角
 {
     return atan2(v.y, v.x);
 }

 bool PointInCircle(Point p, Circle C)       //判断点是否在圆内
 {
     double dist = Length(p - C.c);
     if(dcmp(dist - C.r) > ) return ;      //这里我选择点在圆边上不算在圆内
     else return ;
 }

 bool CircleInCircle(Circle A, Circle B)         //判断圆在圆内
 {
     double cdist = Length(A.c - B.c);
     double rdiff = B.r - A.r;
     if(dcmp(A.r - B.r) <=  && dcmp(cdist - rdiff) <= ) return ;      //包括重合，内切和内含的情况
     return ;
 }

 int n;
 Circle C[maxn];
 bool vis[maxn];
 vector<double> pointAng[maxn];

 int GetCircleCircleIntersection(int c1, int c2)       //求圆与圆的交点
 {
     Circle C1 = C[c1];
     Circle C2 = C[c2];
     double d = Length(C1.c - C2.c);
     if(dcmp(d) == )
     {
         if(dcmp(C1.r - C2.r) == ) return -;       //两圆重合
         return ;       //同心圆但不重合
     }
     if(dcmp(C1.r + C2.r - d) < ) return ;     //外离
     if(dcmp(fabs(C1.r - C2.r) - d) > ) return ;       //内含
     double a = angle(C2.c - C1.c);
     double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / ( * C1.r * d));
     Point p1 = C1.point(a + da);
     Point p2 = C1.point(a - da);
     if(p1 == p2) return ;      //相切
     pointAng[c1].push_back(a + da);     //相切的点不处理，只要相交的
     pointAng[c1].push_back(a - da);
     return ;
 }

 void init()
 {
     for(int i = ; i < n; i++) pointAng[i].clear();
     memset(vis, , sizeof(vis));
 }

 void read()
 {
     for(int i = ; i < n; i++) C[i].read();
 }

 void solve()
 {
     for(int i = ; i < n; i++)      //圆两两相交，得各圆交点集合
         for(int j = ; j < n; j++) if(i != j)
                 GetCircleCircleIntersection(i, j);
     for(int i = ; i < n; i++)
     {
         sort(pointAng[i].begin(), pointAng[i].end());       //各圆交点按极角排序
         vector<double>::iterator iter = unique(pointAng[i].begin(), pointAng[i].end());     //去重，可减少运行时间，不去重也能AC
         pointAng[i].resize(distance(pointAng[i].begin(), iter));
     }
     for(int i = ; i < n; i++)      //判断第i个圆上的弧
     {
         int sz = pointAng[i].size();
         if(!sz)         //此圆不与其他圆相交
         {
             bool ok = ;
             for(int k = i+; k < n; k++) if(CircleInCircle(C[i], C[k]))         //判上面是否有圆把它覆盖掉
                 {
                     ok = ;
                     break;
                 }
             if(ok) vis[i] = ;
         }
         else
         {
             pointAng[i].push_back(pointAng[i][]);
             for(int j = ; j < sz; j++)         //第i个圆上的第j条弧
             {
                 bool ok = ;
                 Point pm = C[i].point((pointAng[i][j] + pointAng[i][j+]) / );     //取弧的中点
                 for(int k = i+; k < n; k++) if(PointInCircle(pm, C[k]))
                     {
                         ok = ;
                         break;
                     }
                 if(ok)
                 {
                     vis[i] = ;
                     for(int u = i-; u >= ; u--)if(PointInCircle(pm, C[u]))        //把这段圆弧下的圆设为可见
                         {
                             vis[u] = ;
                             break;
                         }
                 }
             }
         }
     }
     int ret = ;
     for(int i = ; i < n; i++) if(vis[i]) ret++;
     printf("%d\n", ret);
 }

 int main()
 {
     while(scanf("%d", &n) ==  && n)
     {
         init();
         read();
         solve();
     }
     return ;
 }