Seoul 2007 / UVa 1398 Meteor (排序&线性扫描)

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一场由韩国互联网公司nhn举办的流星雨摄影竞赛中，参赛者通过控制高性能望远镜捕捉尽可能多的流星。为了赢得比赛，选手们需要计算出能够拍摄到最多流星的确切时刻。

1398 - Meteor

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=460&page=show_problem&problem=4144

The famous Korean internet company nhn has provided an internet-based photo service which allows The famous Korean internet company users to directly take a photo of an astronomical phenomenon in space by controlling a high-performance telescope owned by nhn. A few days later, a meteoric shower, known as the biggest one in this century, is expected. nhn has announced a photo competition which awards the user who takes a photo containing as many meteors as possible by using the photo service. For this competition, nhn provides the information on the trajectories of the meteors at their web page in advance. The best way to win is to compute the moment (the time) at which the telescope can catch the maximum number of meteors.

You have n meteors, each moving in uniform linear motion; the meteor m_i moves along the trajectory p_i +t×v_i over time t , where t is a non-negative real value, p_i is the starting point of m_i and v_i is the velocity of m_i . The point p_i = (x_i, y_i) is represented by X -coordinate x_i and Y -coordinate y_i in the (X,Y) -plane, and the velocity v_i = (a_i, b_i) is a non-zero vector with two components a_i and b_i in the (X, Y) -plane. For example, if p_i = (1, 3) and v_i = (-2, 5) , then the meteor m_i will be at the position (0, 5.5) at time t = 0.5 because p_i + t×v_i = (1, 3) + 0.5×(-2, 5) = (0, 5.5) . The telescope has a rectangular frame with the lower-left corner (0, 0) and the upper-right corner (w, h) . Refer to Figure 1. A meteor is said to be in the telescope frame if the meteor is in the interior of the frame (not on the boundary of the frame). For exam! ple, in Figure 1, p₂, p₃, p₄ , and p₅ cannot be taken by the telescope at any time because they do not pass the interior of the frame at all. You need to compute a time at which the number of meteors in the frame of the telescope is maximized, and then output the maximum number of meteors.

$\epsfbox{p3905.eps}$

Input

Your program is to read the input from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing two integers w and h (1 $\le$ w, h $\le$ 100, 000) , the width and height of the telescope frame, which are separated by single space. The second line contains an integer n , the number of input points (meteors), 1 $\le$ n $\le$ 100, 000. Each of the next n lines contain four integers x_i, y_i, a_i , and b_i ; (x_i, y_i) is the starting point p_i and(a_i, b_i) is the nonzero velocity vector v_i of the i -th meteor; x_i and y_i are integer values between -200,000 and 200,000, and a_i and b_i are integer values between -10 and 10. Note that at least one of a_i and b_i is not zero. These four values are separated by single spaces. We assume that all starting points p_i are distinct.

Output

Your program is to write to standard output. Print the maximum number of meteors which can be in the telescope frame at some moment.

Sample Input

Sample Output

1 
2

1. 路径？时间？怎么化繁为简？——对每颗可以被望远镜捕捉到（进入矩形内部）的流星，记录其进入时间和离开时间。

2. 如何计算进出时间？——2个维度分开判断，进入时间取最大值，离开时间取最小值。

3. 怎么判断某一时刻捕捉到的流星数目最多？——排序&扫描：对所有进入和离开的时间进行排序，对于时间相同的点，把离开时间放前面。

4. 扫描？——我把它叫做减增调整法，当有新的扫描事件发生时（即扫描了一个进入时间或离开时间），就减/增流星数目，并比较当前最大值。

完整代码：

/*0.122s*/

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100010;

struct Event
{
	int x;
	int type;
	bool operator < (const Event& a) const
	{
		return x < a.x || (x == a.x && type > a.type); // 先处理离开时间
	}
} events[maxn << 1];

// 0<x+at<w
void update(int x, int a, int w, int& L, int& R)
{
	if (a == 0)
	{
		if (x <= 0 || x >= w) R = L - 1; // 无解
	}
	else if (a > 0)
	{
		L = max(L, -x * 2520 / a); //注意到-10 <= ai,bi <= 10，可以通过乘上2~10的最小公倍数来避免实数运算
		R = min(R, (w - x) * 2520 / a);
	}
	else
	{
		L = max(L, (w - x) * 2520 / a);
		R = min(R, -x * 2520 / a);
	}
}

int main(void)
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int w, h, n, e = 0;
		scanf("%d%d%d", &w, &h, &n);
		for (int i = 0; i < n; i++)
		{
			int x, y, a, b;
			scanf("%d%d%d%d", &x, &y, &a, &b);
			// 0<x+at<w, 0<y+bt<h, t>=0
			int L = 0, R = 1000000000;
			update(x, a, w, L, R);
			update(y, b, h, L, R);//2个维度分开判断
			if (R > L)
			{
				events[e++] = (Event){L, 0};
				events[e++] = (Event){R, 1};
			}
		}
		sort(events, events + e);
		int cnt = 0, ans = 0;
		for (int i = 0; i < e; i++)
		{
			if (events[i].type == 0) ans = max(ans, ++cnt); // 由于要保证ans的取值正确，所以当两时间相同时，把离开时间放前面
			else cnt--;
		}
		printf("%d\n", ans);
	}
	return 0;
}