LA 3905 Meteor 扫描线

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 <tex2html_verbatim_mark>meteors, each moving in uniform linear motion; the meteor m_i <tex2html_verbatim_mark>moves along the trajectory p_i + t×v_i<tex2html_verbatim_mark>over time t <tex2html_verbatim_mark>, where t <tex2html_verbatim_mark>is a non-negative real value, p_i <tex2html_verbatim_mark>is the starting point of m_i <tex2html_verbatim_mark>and v_i <tex2html_verbatim_mark>is the velocity ofm_i <tex2html_verbatim_mark>. The point p_i = (x_i, y_i) <tex2html_verbatim_mark>is represented by X <tex2html_verbatim_mark>-coordinate x_i <tex2html_verbatim_mark>and Y <tex2html_verbatim_mark>-coordinate y_i <tex2html_verbatim_mark>in the (X, Y) <tex2html_verbatim_mark>-plane, and the velocity v_i = (a_i, b_i) <tex2html_verbatim_mark>is a non-zero vector with two components a_i <tex2html_verbatim_mark>and b_i <tex2html_verbatim_mark>in the (X, Y) <tex2html_verbatim_mark>-plane. For example, if p_i = (1, 3) <tex2html_verbatim_mark>and v_i = (-2, 5) <tex2html_verbatim_mark>, then the meteor m_i <tex2html_verbatim_mark>will be at the position (0, 5.5) at time t = 0.5 <tex2html_verbatim_mark>because p_i + t×v_i = (1, 3) + 0.5×(-2, 5) = (0, 5.5) <tex2html_verbatim_mark>. The telescope has a rectangular frame with the lower-left corner (0, 0) and the upper-right corner (w, h) <tex2html_verbatim_mark>. 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 <tex2html_verbatim_mark>, and p₅ <tex2html_verbatim_mark>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.

<tex2html_verbatim_mark>

Input

Your program is to read the input from standard input. The input consists of T <tex2html_verbatim_mark>test cases. The number of test cases T <tex2html_verbatim_mark>is given in the first line of the input. Each test case starts with a line containing two integers w<tex2html_verbatim_mark>and h <tex2html_verbatim_mark>(1w, h100, 000) <tex2html_verbatim_mark>, the width and height of the telescope frame, which are separated by single space. The second line contains an integer n <tex2html_verbatim_mark>, the number of input points (meteors), 1n100, 000 <tex2html_verbatim_mark>. Each of the next n <tex2html_verbatim_mark>lines contain four integers x_i, y_i, a_i <tex2html_verbatim_mark>, and b_i <tex2html_verbatim_mark>; (x_i, y_i) <tex2html_verbatim_mark>is the starting point p_i <tex2html_verbatim_mark>and (a_i, b_i) <tex2html_verbatim_mark>is the nonzero velocity vector v_i <tex2html_verbatim_mark>of the i <tex2html_verbatim_mark>-th meteor; x_i <tex2html_verbatim_mark>and y_i <tex2html_verbatim_mark>are integer values between -200,000 and 200,000, and a_i <tex2html_verbatim_mark>and b_i <tex2html_verbatim_mark>are integer values between -10 and 10. Note that at least one of a_i <tex2html_verbatim_mark>and b_i <tex2html_verbatim_mark>is not zero. These four values are separated by single spaces. We assume that all starting points p_i <tex2html_verbatim_mark>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

题目大意：XOY坐标系上，给n个点(每个点的起始坐标跟速度)，求在矩形区域最多能同时出现多少个点。

先把每个点在矩形上出现的时间段求出来（左右端点（左标记为0，右标记为1）push进vector容器，），二级排序。

从左只有处理各个端点，每遇到一个左端点，计数器加一；每遇到一个右端点，计数器减一。每次更新最大值。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <algorithm>
 using namespace std;

 const double eps=1e-;
 double max(double a,double b){ return a-b>eps?a:b;}
 double min(double a,double b){ return b-a>eps?a:b;}

 struct node
 {
     double x;
     int i;
     node(double x=,int i=):x(x),i(i){}
     bool operator<(const node &A)const{
         if(fabs(A.x-x)>eps) return x<A.x;
         else return i>A.i;
     }
 };
 int w,h,n;
 vector<node> v;

 void update(int x,int a,int w,double &L,double &R)
 {
     if(a==)
     {
         if(x<= || x>=w) R=L-;
     }
     else if(a>)
     {
         L=max(L,-(double)x/a);
         R=min(R,(double)(w-x)/a);
     }
     else
     {
         L=max(L,(double)(w-x)/a);
         R=min(R,-(double)x/a);
     }
 }

 void Push()
 {
     int x,y,a,b;
     scanf("%d %d %d %d",&x,&y,&a,&b);
     double L=,R=1e9;
     update(x,a,w,L,R);
     update(y,b,h,L,R);
     if(R-L>eps)//经过矩形的起止时间
     {
         v.push_back(node(L,));
         v.push_back(node(R,));
     }
 }

 int main()
 {
     int T,i;
     scanf("%d",&T);
     while(T--)
     {
         v.clear();
         scanf("%d %d %d",&w,&h,&n);
         for(i=;i<n;i++) Push();
         sort(v.begin(),v.end());
         int cnt=,ans=;
         for(i=;i<v.size();i++)
         {
             if(v[i].i==) ans=max(ans,++cnt);
             else cnt--;
         }
         printf("%d\n",ans);
     }
     return ;
 }