poj1696

寻找逆时针螺旋线最多能连几个点，每次都连最外面的点即可。极角排序。换做max_element可提高效率，但这个题规模太小了，没必要。综合复杂度(n^2)log(n)。

请结合图片理解此过程：

#include
#include
#include

using namespace std;

struct Point
{
	int x, y, num;
};

Point p[55], res[55];
int pos;

int cross(Point p0, Point p1, Point p2)
{
	return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
}

int Distance(Point p1, Point p2)
{
	return (p2.x - p1.x)*(p2.x - p1.x) + (p2.y - p1.y)*(p2.y - p1.y);
}

bool cmp(const Point & p1, const Point & p2)
{
	int temp;
	temp = cross(p[pos], p1, p2);
	if (temp > 0)
	{
		return true;
	}
	else if (temp == 0 && Distance(p[pos], p1) < Distance(p[pos], p2))
	{
		return true;
	}
	return false;
}

int main()
{
	int ncase;
	cin >> ncase;
	while (ncase--)
	{
		int n;
		cin >> n;
		for (int i = 0; i < n; i++)
		{
			cin>>p[i].num>>p[i].x>>p[i].y;
			if (p[i].y < p[0].y)
			{
				swap(p[0], p[i]);
			}
		}

		int j = pos = 0;

		sort(p + 1, p + n, cmp);
		res[j++] = p[pos++];

		for (int i = 2; i < n; i++)
		{
			sort(p + pos, p + n, cmp);
			res[j++] = p[pos++];
		}

		res[j++] = p[pos++];
		printf("%d", j);

		for (int i = 0; i < j; i++)
		{
			printf(" %d", res[i].num);
		}
		printf("\n");
	}
}