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 ;
}