[半平面交] poj 3384 Feng Shui

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

const double eps = 1e-8;

inline double delt(double a)
{
    return (fabs(a) < eps ? 0 : a) > 0 ? 1 : -1;
}

#define N 1005
#define inf 1e20

using namespace std;

struct TPoint
{
    double x, y;
};

inline double dist(TPoint a, TPoint b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

inline double cross(TPoint a, TPoint b, TPoint c)
{
    return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}

inline double dot(TPoint a, TPoint b, TPoint c)
{
    return (b.x - a.x) * (c.x - a.x) + (b.y - a.y) * (c.y - a.y);
}

TPoint pt[N], st, mid[N];
int n, m, t, end;
double radius;

bool scan()
{
	for (int i = 0; i < n; i++)
	{
		scanf("%lf%lf", &pt[i].x, &pt[i].y);
	}
	return 1;
}

inline bool cmp(TPoint a, TPoint b)
{
	return a.x < b.x || (a.x == b.x && a.y < b.y);
}

inline bool ncmp(TPoint a, TPoint b)
{
	int d1 = delt(cross(pt[0], a, b));
	return d1 > 0 || (d1 == 0 && dist(pt[0], a) < dist(pt[0], b));
}

void getmove(TPoint &a, TPoint &b, double r)
{
	TPoint ms;
	ms.x = a.y - b.y, ms.y = b.x - a.x;
	double k = r / dist(a, b);
	a.x += ms.x * k, a.y += ms.y * k;
	b.x += ms.x * k, b.y += ms.y * k;
}

void cut(TPoint ans[], TPoint s, TPoint e, int &np)
{
	getmove(s, e, radius);
	double s1, s2;
	int i, j, d1, d2;
	for (i = j = 0; i < np; i++)
	{
		s1 = cross(ans[i], s, e), s2 = cross(ans[i + 1], s, e);
		d1 = delt(s1), d2 = delt(s2);
		if (d1 >= 0) ans[j++] = ans[i];
		if (d1 * d2 < 0)
		{
			ans[j].x = (s2 * ans[i].x - s1 * ans[i + 1].x) / (s2 - s1);
			ans[j++].y = (s2 * ans[i].y - s1 * ans[i + 1].y) / (s2 - s1);
		}
	}
	ans[j] = ans[0];
	np = j;
}

void solve(void)
{
	sort(pt, pt + n, cmp);
	sort(pt, pt + n, ncmp);
	int i, j, k, maxl = 0;
	TPoint ans[N], a, b;
	pt[n] = pt[0];
	for (i = 0; i <= n; i++) ans[i] = pt[i];
	for (i = 0, m = n; i < n; i++) cut(ans, pt[i], pt[i + 1], m);

	if (m == 1) a = b = ans[0];
	else if (m == 2) a = ans[0], b = ans[1];
	else
	{
		double maxl = 0, l;
		for (i = 0; i < m; i++)
		{
			for (j = 0; j < m; j++)
			{
				l = dist(ans[i], ans[j]);
				if (l - maxl > eps)
				{
					maxl = l, a = ans[i], b = ans[j];
				}
			}
		}
	}
	printf("%.4f %.4f %.4f %.4f\n", a.x, a.y, b.x, b.y);
}

int main(void)
{
    while(scanf("%d%lf", &n, &radius) != EOF)
    {
		scan();
		solve();
    }
    return 0;
}

题意：给出一个凸多边形的房间，根据风水要求，把两个圆形地毯铺在房间里，不能折叠，不能切割，可以重叠。问最多能覆盖多大空间，输出两个地毯的圆心坐标。多组解输出其中一个

题目保证至少可以放入一个圆，上一题中判断过在一个多边形内是否能放入一个半径为r的圆。同样将多边形的边内移R之后，半平面交区域便是可以放入圆的可行区域。题目要求覆盖的面积最大，也就是两个圆的半径相同，圆心越远，面积就越大。在半平面交区域内找到最远点对便是题目要求的解。