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 = 110;
const double PI = acos(-1.0);
const double Two_PI = PI * 2;
const double eps = 5 * 1e-13;
int n;
double radius[maxn];
bool vis[maxn];

int dcmp(double x)
{
    if(fabs(x) < eps)    return 0;
    else return x < 0 ? -1 : 1;
}

struct Point
{
    double x, y;
    Point(double x=0, double y=0):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) == 0)    return;
    if(dcmp(d-r1-r2) > 0)    return;
    if(dcmp(d-fabs(r1-r2)) < 0)    return;
    
    double a = angle(c2-c1);
    double ag = acos((r1*r1+d*d-r2*r2)/(2*r1*d));
    rad.push_back(NormalizeAngle(a + ag));
    rad.push_back(NormalizeAngle(a - ag));
}

int topmost(Point P)
{
    for(int i = n-1; i >= 0; --i)
        if(Length(P-center[i]) < radius[i])    return i;
    return -1;
}

int main(void)
{
    #ifdef LOCAL
        freopen("2572in.txt", "r", stdin);
    #endif
    
    while(scanf("%d", &n) == 1 && n)
    {
        for(int i = 0; i < n; ++i)
            scanf("%lf%lf%lf", &center[i].x, &center[i].y, &radius[i]);
        memset(vis, 0, sizeof(vis));
        
        for(int i = 0; i < n; ++i)
        {
            vector<double> rad;
            rad.push_back(0.0);
            rad.push_back(Two_PI);
            for(int j = 0; j < n && j != i; ++j)
                getCCIntersection(center[i], radius[i], center[j], radius[j], rad);
            sort(rad.begin(), rad.end());
            
            for(int j = 0; j < rad.size() - 1; ++j)
            {
                double mid = (rad[j] + rad[j+1]) / 2.0;
                for(int side = -1; side <= 1; side += 2)
                {
                    double r = radius[i] + side*eps;
                    int t = topmost(Point(center[i].x+r*cos(mid), center[i].y+r*sin(mid)));
                    if(t >= 0)    vis[t] = true;
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < n; ++i)    if(vis[i])    ++ans;
        printf("%d
", ans);
    }

    return 0;
}