ZOJ-2102

比较巧妙的物理几何题，参考自大神的解题报告，题意就是给一些不重复的圆，然后把一根棍子放到这些圆上，判断会不会掉下来，在物理上来说就是重心（即线段中心）两端都要有点在圆上，这样才能保证不掉下来，思路就是把线段均分，依次枚举所有切分点，判断这些点是否在圆上，然后看看最外侧的两个在圆上的点是否在重心两侧

#include<cstdio>
#include<vector>

using namespace std;

namespace
{
	const int D = 500;
	const double eps = 1e-8;

	struct Circle
	{
		double x, y, r;
	};

	int sgn(double a)
	{
		return a < -eps ? -1 : a < eps ? 0 : 1;
	}

	bool in_circle(double x, double y, vector<Circle *> &V)
	{
		for (size_t i = 0; i < V.size(); i++)
		{
			double dis = (x - V[i]->x) * (x - V[i]->x)
					+ (y - V[i]->y) * (y - V[i]->y) - (V[i]->r) * (V[i]->r);
			if (sgn(dis) <= 0)
				return true;
		}
		return false;
	}
}

int main()
{
	int N;
	vector<Circle *> V;
	vector<pair<double, double> > U;
	double x, y, bx, by, ex, ey;
	while (scanf("%d", &N), N)
	{
		V.clear();
		U.clear();
		while (N--)
		{
			Circle *c = new Circle();
			scanf("%lf %lf %lf", &(c->x), &(c->y), &(c->r));
			V.push_back(c);
		}
		scanf("%lf %lf %lf %lf", &bx, &by, &ex, &ey);
		double sx = (ex - bx) / D, sy = (ey - by) / D;
		if (in_circle(bx, by, V))
			U.push_back(make_pair(bx, by));
		for (int i = 1; i < D; i++)
		{
			x = bx + i * sx;
			y = by + i * sy;
			if (in_circle(x, y, V))
				U.push_back(make_pair(x, y));
		}
		if (in_circle(ex, ey, V))
			U.push_back(make_pair(ex, ey));

		if (U.size() >= 2)
		{
			double mx = (bx + ex) / 2, my = (by + ey) / 2;
			double fx = U.front().first, fy = U.front().second;
			double lx = U.back().first, ly = U.back().second;

			if ((fx - mx) * (lx - mx) + (fy - my) * (ly - my) < 0)
				puts("STAY");
			else
				puts("FALL");
		}
		else
			puts("FALL");

		for (size_t i = 0; i < V.size(); i++)
			delete V[i];
	}
	return 0;
}