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.html
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.node
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 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.ios
Your program is to write to standard output. Print the maximum number of meteors which can be in the telescope frame at some moment.web
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
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題目大意:XOY座標系上,給n個點(每一個點的起始座標跟速度),求在矩形區域最多能同時出現多少個點。
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先把每一個點在矩形上出現的時間段求出來(左右端點(左標記爲0,右標記爲1)push進vector容器,),二級排序。code
從左只有處理各個端點,每遇到一個左端點,計數器加一;每遇到一個右端點,計數器減一。每次更新最大值。component
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)//通過矩形的起止時間 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 }