题意：N(1<=N<=20) 个圆，现取其中一个圆的圆心为圆心作一个大圆，使得这个大圆与其他任意一个圆 c 的面积交 >= c 的面积的一半，求大圆的最小半径。
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3264
——>>枚举圆心，二分半径。。
#include <cstdio>
#include <cmath>
#include <algorithm>using std::min;const double EPS = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 20 + 10;int N;
double harea[MAXN];int Dcmp(double x)
{if (fabs(x) < EPS) return 0;return x > 0 ? 1 : -1;
}struct POINT
{double x;double y;POINT(double x = 0.0, double y = 0.0) : x(x), y(y) {}
};
typedef POINT VECTOR;struct CIRCLE
{POINT c;double r;CIRCLE() {}CIRCLE(POINT c, double r) : c(c), r(r) {}
} mall[MAXN];double Length(POINT A, POINT B)
{return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}// 两圆面积交
double AreaCircleCircle(CIRCLE c1, CIRCLE c2)
{double ret = 0.0;double d = Length(c1.c, c2.c);if (Dcmp(d - c1.r - c2.r) < 0 && Dcmp(d - fabs(c1.r - c2.r)) > 0){double a1 = acos((c1.r * c1.r + d * d - c2.r * c2.r) / 2 / c1.r / d);double a2 = acos((c2.r * c2.r + d * d - c1.r * c1.r) / 2 / c2.r / d);ret = (a1 - sin(2 * a1) / 2) * c1.r * c1.r + (a2 - sin(2 * a2) / 2) * c2.r * c2.r;}else if (Dcmp(d - fabs(c1.r - c2.r)) <= 0){if (Dcmp(c1.r - c2.r) < 0){ret = PI * c1.r * c1.r;}else{ret = PI * c2.r * c2.r;}}return ret;
}void Read()
{scanf("%d", &N);for (int i = 0; i < N; ++i){scanf("%lf%lf%lf", &mall[i].c.x, &mall[i].c.y, &mall[i].r);harea[i] = PI * mall[i].r * mall[i].r / 2;}
}bool IsOk(const POINT& center, const double& r)
{for (int i = 0; i < N; ++i){if (Dcmp(AreaCircleCircle(CIRCLE(center, r), mall[i]) - harea[i]) < 0){return false;}}return true;
}void Solve()
{double ret = 34000.0;for (int i = 0; i < N; ++i){double L = 0.0;double R = 34000.0;while (Dcmp(L - R) < 0){double M = (L + R) / 2;IsOk(mall[i].c, M) ? R = M : L = M;}ret = min(ret, L);}printf("%.4f\n", ret);
}int main()
{int T;scanf("%d", &T);while (T--){Read();Solve();}return 0;
}