LA 2572 (求可见圆盘的数量) Kanazawa

题意：

把n个圆盘依次放到桌面上，按照放置的先后顺序给出这n个圆盘的圆心和半径，输出有多少个圆盘可见(即未被全部覆盖)。

分析：

题中说对输入数据进行微小扰动后答案不变。

所露出的部分都是由若干小圆弧组成的。因此求出每个圆与其他圆相交的小圆弧，取圆弧的终点，分别向里和向外移动一个很小的距离的到P'。标记包含P'的最上面的那个圆为可见。

 //#define LOCAL
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <vector>
 using namespace std;

 const int maxn = ;
 const double PI = acos(-1.0);
 const double Two_PI = PI * ;
 const double eps =  * 1e-;
 int n;
 double radius[maxn];
 bool vis[maxn];

 int 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;
 Point operator + (Point A, Point B)    { return Point(A.x+B.x, A.y+B.y); }
 Point operator - (Point A, Point B)    { return Point(A.x-B.x, A.y-B.y); }
 Vector operator * (Vector A, double p){ return Vector(A.x*p, A.y*p); }
 Vector operator / (Vector A, double p){ return Vector(A.x/p, A.y/p); }
 Point center[maxn];

 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)    { return atan2(v.y, v.x); }
 double NormalizeAngle(double rad)
 {//将所求角度化到0到2π之间
     return rad - Two_PI*(floor(rad/Two_PI));
 }

 void getCCIntersection(Point c1, double r1, Point c2, double r2, vector<double>& rad)
 {
     double d = Length(c1 - c2);
     if(dcmp(d) == )    return;
     if(dcmp(d-r1-r2) > )    return;
     if(dcmp(d-fabs(r1-r2)) < )    return;

     double a = angle(c2-c1);
     double ag = acos((r1*r1+d*d-r2*r2)/(*r1*d));
     rad.push_back(NormalizeAngle(a + ag));
     rad.push_back(NormalizeAngle(a - ag));
 }

 int topmost(Point P)
 {
     for(int i = n-; i >= ; --i)
         if(Length(P-center[i]) < radius[i])    return i;
     return -;
 }

 int main(void)
 {
     #ifdef LOCAL
         freopen("2572in.txt", "r", stdin);
     #endif

     while(scanf("%d", &n) ==  && n)
     {
         for(int i = ; i < n; ++i)
             scanf("%lf%lf%lf", &center[i].x, &center[i].y, &radius[i]);
         memset(vis, , sizeof(vis));

         for(int i = ; i < n; ++i)
         {
             vector<double> rad;
             rad.push_back(0.0);
             rad.push_back(Two_PI);
             for(int j = ; j < n && j != i; ++j)
                 getCCIntersection(center[i], radius[i], center[j], radius[j], rad);
             sort(rad.begin(), rad.end());

             for(int j = ; j < rad.size() - ; ++j)
             {
                 double mid = (rad[j] + rad[j+]) / 2.0;
                 for(int side = -; side <= ; side += )
                 {
                     double r = radius[i] + side*eps;
                     int t = topmost(Point(center[i].x+r*cos(mid), center[i].y+r*sin(mid)));
                     if(t >= )    vis[t] = true;
                 }
             }
         }
         int ans = ;
         for(int i = ; i < n; ++i)    if(vis[i])    ++ans;
         printf("%d\n", ans);
     }

     return ;
 }

代码君