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 mi <tex2html_verbatim_mark>moves
along the trajectory pi + t×vi<tex2html_verbatim_mark>over
time t <tex2html_verbatim_mark>,
where t <tex2html_verbatim_mark>is a
non-negative real value, pi <tex2html_verbatim_mark>is
the starting point of mi <tex2html_verbatim_mark>and vi <tex2html_verbatim_mark>is
the velocity ofmi <tex2html_verbatim_mark>.
The point pi =
(xi, yi) <tex2html_verbatim_mark>is
represented by X <tex2html_verbatim_mark>-coordinate xi <tex2html_verbatim_mark>and Y <tex2html_verbatim_mark>-coordinate yi <tex2html_verbatim_mark>in
the (X, Y) <tex2html_verbatim_mark>-plane,
and the velocity vi =
(ai, bi) <tex2html_verbatim_mark>is
a non-zero vector with two components ai <tex2html_verbatim_mark>and bi <tex2html_verbatim_mark>in
the (X, Y) <tex2html_verbatim_mark>-plane.
For example, if pi = (1,
3) <tex2html_verbatim_mark>and vi = (-2,
5) <tex2html_verbatim_mark>, then the meteor mi <tex2html_verbatim_mark>will
be at the position (0, 5.5) at time t =
0.5 <tex2html_verbatim_mark>because pi + t×vi =
(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>(1
w, 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 xi, yi, ai <tex2html_verbatim_mark>,
and bi <tex2html_verbatim_mark>; (xi, yi) <tex2html_verbatim_mark>is
the starting point pi <tex2html_verbatim_mark>and (ai, bi) <tex2html_verbatim_mark>is
the nonzero velocity vector vi <tex2html_verbatim_mark>of
the i <tex2html_verbatim_mark>-th
meteor; xi <tex2html_verbatim_mark>and yi <tex2html_verbatim_mark>are
integer values between -200,000 and 200,000, and ai <tex2html_verbatim_mark>and bi <tex2html_verbatim_mark>are
integer values between -10 and 10. Note that at least one of ai <tex2html_verbatim_mark>and bi <tex2html_verbatim_mark>is
not zero. These four values are separated by single spaces. We assume that all
starting points pi <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。
1 #include <iostream>
2 #include <cstdio>
3 #include <cmath>
4 #include <vector>
5 #include <algorithm>
6 using namespace std;
7
8 const double eps=1e-9;
9 double max(double a,double b){ return a-b>eps?a:b;}
10 double min(double a,double b){ return b-a>eps?a:b;}
11
12 struct node
13 {
14 double x;
15 int i;
16 node(double x=0,int i=0):x(x),i(i){}
17 bool operator<(const node &A)const{
18 if(fabs(A.x-x)>eps) return x<A.x;
19 else return i>A.i;
20 }
21 };
22 int w,h,n;
23 vector<node> v;
24
25 void update(int x,int a,int w,double &L,double &R)
26 {
27 if(a==0)
28 {
29 if(x<=0 || x>=w) R=L-1;
30 }
31 else if(a>0)
32 {
33 L=max(L,-(double)x/a);
34 R=min(R,(double)(w-x)/a);
35 }
36 else
37 {
38 L=max(L,(double)(w-x)/a);
39 R=min(R,-(double)x/a);
40 }
41 }
42
43 void Push()
44 {
45 int x,y,a,b;
46 scanf("%d %d %d %d",&x,&y,&a,&b);
47 double L=0,R=1e9;
48 update(x,a,w,L,R);
49 update(y,b,h,L,R);
50 if(R-L>eps)//经过矩形的起止时间
51 {
52 v.push_back(node(L,0));
53 v.push_back(node(R,1));
54 }
55 }
56
57 int main()
58 {
59 int T,i;
60 scanf("%d",&T);
61 while(T--)
62 {
63 v.clear();
64 scanf("%d %d %d",&w,&h,&n);
65 for(i=0;i<n;i++) Push();
66 sort(v.begin(),v.end());
67 int cnt=0,ans=0;
68 for(i=0;i<v.size();i++)
69 {
70 if(v[i].i==0) ans=max(ans,++cnt);
71 else cnt--;
72 }
73 printf("%d\n",ans);
74 }
75 return 0;
76 }