杭电OJ 1109（C++）

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int MAXN = 1505;
const int NUM = 30; //随机选择的初始点的数量
const double exps = 1e-3; //精度
const double pi = acos(-1.0); //圆周率Π

double PX[MAXN], PY[MAXN]; //陷阱的X坐标和Y坐标
double fx[MAXN], fy[MAXN], best[MAXN]; //随机选择的点的坐标和到达最近陷阱的距离

//在区间内生成随机数
double Rand(double L, double R)
{
	return (rand() % 10000) / 10000.0 * (R - L) + L;
}

//计算两点间的距离
double dist(double x1, double y1, double x2, double y2)
{
	return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

int main()
{
	int T, X, Y, n;
	cin >> T;
	srand(time(0));
	while (T--)
	{
		cin >> X >> Y >> n;
		for (int i = 1; i <= n; i++)
		{
			cin >> PX[i] >> PY[i];
		}

		for (int i = 1; i <= NUM; i++) //随机NUM个初始点，进行模拟退火
		{
			fx[i] = Rand(1, X);
			fy[i] = Rand(1, Y);
			best[i] = INT_MAX;
			for (int j = 1; j <= n; j++)
			{
				//评估函数，评估整个退火过程的好坏
				best[i] = min(best[i], dist(fx[i], fy[i], PX[j], PY[j]));
			}
		}

		double step = max(X, Y); //最长跨度
		while (step > exps)
		{
			for (int i = 1; i <= NUM; i++)
			{
				for (int j = 1; j <= NUM; j++)
				{
					double angel = Rand(0, 2 * pi); //枚举任意角度，使得到的新点向四周扩散
					double nx = fx[i] + cos(angel) * step;
					double ny = fy[i] + sin(angel) * step;
					if (nx < 0 || nx > X || ny < 0 || ny > Y)
						continue;

					double d = INT_MAX;
					for (int k = 1; k <= n; k++)
					{
						d = min(d, dist(nx, ny, PX[k], PY[k]));
					}
					if (d > best[i])
					{
						best[i] = d;
						fx[i] = nx;
						fy[i] = ny;
					}
				}
			}
			step *= 0.85; //退火
		}
		int t = 1;
		for (int i = 1; i <= NUM; i++)
		{
			if (best[i] >= best[t])
				t = i;
		}
		cout << "The safest point is (" 
			 << fixed << setprecision(1) << fx[t] << ", " 
			 << fy[t] << ")." << endl;
	}
	return 0;
}