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