uvalive3905（掃描法）

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 meteorm_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 ofm_i. The point p_i = (x_i,y_i) is represented by X-coordinatex_i and Y-coordinatey_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 meteorm_i will be at the position (0, 5.5) at timet = 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, andp₅ 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.

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 integersw and h (1w,h100, 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),1n100, 000. Each of the next n lines contain four integersx_i, y_i, a_i, andb_i; (x_i,y_i) is the starting point p_i and(a_i, b_i) is the nonzero velocity vectorv_i of the i-th meteor;x_i and y_i are integer values between -200,000 and 200,000, anda_i and b_i are integer values between -10 and 10. Note that at least one ofa_i and b_i is not zero. These four values are separated by single spaces. We assume that all starting pointsp_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 

2 
4 2 
2 
-1 1 1 -1 
5 2 -1 -1 
13 6 
7 
3 -2 1 3 
6 9 -2 -1 
8 0 -1 -1 
7 6 10 0 
11 -2 2 1 
-2 4 6 -1 
3 2 -5 -1

Sample Output 

1 
2

劉汝佳大白書上的掃描法

將有效的流星進入攝像頭和離開攝像頭的時間處理爲一個區間，然後把區間左右端點處理成兩個事件，遇到左事件答案加1，遇到右事件答案減一，取最大就是最後的答案，我是一個渣渣，所以代碼跟劉汝佳的差不多，弱爆了！！！！！！！

#include<set>
#include<queue>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<utility>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define Inf (1<<30)
#define LL long long
#define MOD 1000000009
#pragma comment(linker, "/STACK:102400000,102400000")  
using namespace std;
const int MM = 100005;
struct node
{
	double x;
	int type;
	bool operator <(const node &other)const{
		if (x==other.x)return type < other.type;//右端點是0，左端點是1
		return x < other.x;
	}
}A[MM*2];
void update(int x, int v, int w, int &L, int &R)
{
	if (v == 0)
	{
		if (x <= 0 || x >= w)R = L - 1;
	}
	else if (v>0)
	{
		L = max(L, -x * 2520 / v);
		R = min(R, (w - x) * 2520 / v);
	}
	else
	{
		L = max(L, (w - x) * 2520 / v);
		R = min(R, -x * 2520 / v);
	}
}
int main() {
	int T, n, w, h;
	//freopen("D:\\oo.txt","r",stdin);
	scanf("%d", &T);
	while (T--)
	{
		int cnt = 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);
			int L = 0, R = 1e9;
			update(x, a, w, L, R);
			update(y, b, h, L, R);
			if (R>L)
			{
				A[cnt++] = node{ L, 1 };
				A[cnt++] = node{ R, 0 };
			}
		}
		sort(A, A + cnt);
		int num = 0, ans = 0;
		for (int i = 0; i < cnt; i++)
		{
			if (A[i].type == 1)num++;
			else num--;
			ans = max(num, ans);
		}
		printf("%d\n", ans);
	}
	return 0;
}