uva 10969

题意：平面上依次放置n个圆，后放的覆盖先放的，按顺序给出每个圆的半径和圆心坐标，问最后图形的可见圆弧长之和。
题解：因为是后放的覆盖先放的，所以逆序枚举，每个圆只考虑之前放过的圆和自己的交点，把所有交点排序后可以得到每两个相邻的交点之间的圆弧，找到圆弧中点，如果这个点在之前放过的圆内，说明这个圆弧不能要，否则加到答案里。注意弧度要规范到0～2π。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-11;
struct Point {
    double x, y;
    Point(double x = 0, double y = 0): x(x), y(y) {}
};
typedef Point Vector;

struct Circle {
    Point c;
    double r;
    Circle() {}
    Circle(Point c, double r = 0): c(c), r(r) {}
    Point point(double a) {
        return Point(c.x + cos(a) * r, c.y + sin(a) * c.y);
    }
};

double dcmp(double x) {
    if (fabs(x) < eps)
        return 0;
    return x < 0 ? -1 : 1;
}
Vector operator + (const Point& A, const Point& B) {
    return Vector(A.x + B.x, A.y + B.y);
}
Vector operator - (const Point& A, const Point& B) {
    return Vector(A.x - B.x, A.y - B.y);
}
Vector operator * (const Point& A, double a) {
    return Vector(A.x * a, A.y * a);
}
Vector operator / (const Point& A, double a) {
    return Vector(A.x / a, A.y / a);
}
double Cross(const Vector& A, const Vector& B) {
    return A.x * B.y - A.y * B.x;
}
double Dot(const Vector& A, const Vector& B) {
    return A.x * B.x + A.y * B.y;
}
double Length(const Vector& A) {
    return sqrt(Dot(A, A));
}
double Angle(Vector A, Vector B) {
    return acos(Dot(A, B) / Length(A) / Length(B));
}
bool operator < (const Point& A, const Point& B) {
    return A.x < B.x || (A.x == B.x && A.y < B.y);
}
bool operator == (const Point& A, const Point& B) {
    return A.x == B.x && A.y == B.y;
}
Point Rotate(Point A, double rad) {
    return Point(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}
double torad(double deg) {
    return deg / 180.0 * PI;
}
double normal(double rad) {
    return rad - 2.0 * M_PI * floor(rad / 2.0 / M_PI);
}
void GetCircleIntersection(Circle a, Circle b, vector<double>& v) {
    if (dcmp(Length(a.c - b.c)) == 0)
        return;
    if (dcmp(a.r + b.r - Length(a.c - b.c)) < 0)
        return;
    if (dcmp(fabs(a.r - b.r) - Length(a.c - b.c)) > 0)
        return;
    double rad = atan2(b.c.y - a.c.y, b.c.x - a.c.x);
    double rad1 = acos((a.r * a.r + Length(a.c - b.c) * Length(a.c - b.c) - b.r * b.r) / (2.0 * a.r * Length(a.c - b.c)));
    v.push_back(normal(rad - rad1));
    v.push_back(normal(rad + rad1));
}

const int N = 105;
int n;
Circle cir[N];

bool judge(int cur, Point p) {
    for (int i = n - 1; i > cur; i--)
        if (dcmp(Length(cir[i].c - p) - cir[i].r) <= 0)
            return false;
    return true;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
            scanf("%lf%lf%lf", &cir[i].r, &cir[i].c.x, &cir[i].c.y);
        double res = 0;
        for (int i = n - 1; i >= 0; i--) {
            vector<double> v;
            v.push_back(0);
            v.push_back(PI * 2);
            for (int j = n - 1; j > i; j--)
                GetCircleIntersection(cir[i], cir[j], v);
            sort(v.begin(), v.end());
            double temp = 0;
            for (int j = 0; j < v.size() - 1; j++) {
                double mid = (v[j] + v[j + 1]) / 2.0;
                if (judge(i, Point(cir[i].c.x + cos(mid) * cir[i].r, cir[i].c.y + sin(mid) * cir[i].r)))
                    temp += v[j + 1] - v[j];
            }
            res += temp * cir[i].r;
        }
        printf("%.3lf\n", res);
    }
    return 0;
}