LA - 2572 - Viva Confetti

题意：先后给出n个圆，问最后能看到多少个圆(n <= 100，输入数据小数位数可达12位)。

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=573

——>>圆两两相交，得到交点（保存其极角就行），每个圆内交点的极角排序，扫描每一小段弧，判断其弧中点（不是弦中点）是否在上面的圆内，若这段小弧可见，再将其能看到的下面的第一个圆设为可见。

注意：精度eps不能太大。

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

using namespace std;

const int maxn = 100 + 10;
const double eps = 1e-14;        //别开太大，样例数据就到达1e-11级别
const double pi = acos(-1);

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

struct Point{
    double x;
    double y;

    Point(double x = 0, double y = 0):x(x), y(y){}

    bool operator < (const Point& e) const{
        return dcmp(x - e.x) < 0 || (dcmp(x - e.x) == 0 && dcmp(y - e.y) < 0);
    }

    bool operator == (const Point& e) const{
        return dcmp(x - e.x) == 0 && dcmp(y - e.y) == 0;
    }

    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) > 0) return 0;      //这里我选择点在圆边上不算在圆内
    else return 1;
}

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) <= 0 && dcmp(cdist - rdiff) <= 0) return 1;      //包括重合，内切和内含的情况
    return 0;
}

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) == 0){
        if(dcmp(C1.r - C2.r) == 0) return -1;       //两圆重合
        return 0;       //同心圆但不重合
    }
    if(dcmp(C1.r + C2.r - d) < 0) return 0;     //外离
    if(dcmp(fabs(C1.r - C2.r) - d) > 0) return 0;       //内含
    double a = angle(C2.c - C1.c);
    double da = acos((C1.r * C1.r + d * d - C2.r * C2.r) / (2 * C1.r * d));
    Point p1 = C1.point(a + da);
    Point p2 = C1.point(a - da);
    if(p1 == p2) return 1;      //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 2;
}

void init(){
    for(int i = 0; i < n; i++) pointAng[i].clear();
    memset(vis, 0, sizeof(vis));
}

void read(){
    for(int i = 0; i < n; i++) C[i].read();
}

void solve(){
    for(int i = 0; i < n; i++)      //圆两两相交，得各圆交点集合
        for(int j = 0; j < n; j++) if(i != j)
            GetCircleCircleIntersection(i, j);
    for(int i = 0; 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 = 0; 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 = 1;
            for(int k = i+1; k < n; k++) if(CircleInCircle(C[i], C[k])){        //判上面是否有圆把它覆盖掉
                ok = 0;
                break;
            }
            if(ok) vis[i] = 1;
        }
        else{
            pointAng[i].push_back(pointAng[i][0]);
            for(int j = 0; j < sz; j++){        //第i个圆上的第j条弧
                bool ok = 1;
                Point pm = C[i].point((pointAng[i][j] + pointAng[i][j+1]) / 2);     //取弧的中点
                for(int k = i+1; k < n; k++) if(PointInCircle(pm, C[k])){
                    ok = 0;
                    break;
                }
                if(ok){
                    vis[i] = 1;
                    for(int u = i-1; u >= 0; u--)if(PointInCircle(pm, C[u])){       //把这段圆弧下的圆设为可见
                        vis[u] = 1;
                        break;
                    }
                }
            }
        }
    }
    int ret = 0;
    for(int i = 0; i < n; i++) if(vis[i]) ret++;
    printf("%d\n", ret);
}

int main()
{   //freopen("data.txt", "r", stdin);      //提交时忘了注释掉这个吃了2个WA。。。
    while(scanf("%d", &n) == 1 && n){
        init();
        read();
        solve();
    }
    return 0;
}