3905 - Meteor

链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1906

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 trajectoryp_i + t×v_i over timet , where t is a non-negative real value,p_i is the starting point ofm_i and v_i is the velocity ofm_i . The pointp_i = (x_i,y_i) is represented byX -coordinatex_i andY -coordinatey_i in the(X,Y) -plane, and the velocity v_i = (a_i,b_i) is a non-zero vector with two componentsa_i and b_i in the(X,Y) -plane. For example, if p_i = (1, 3) andv_i = (-2, 5) , then the meteorm_i will be at the position (0, 5.5) at timet = 0.5 becausep_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 andh (1w,h100, 000) , the width and height of the telescope frame, which are separated by single space. The second line contains an integern , 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 pointp_i and(a_i,b_i) is the nonzero velocity vectorv_i of thei -th meteor;x_i andy_i are integer values between -200,000 and 200,000, anda_i andb_i are integer values between -10 and 10. Note that at least one ofa_i andb_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

分析：

流星的轨迹是没有意义的，有用的只是每一点出现的时间，type=0，1只是左右端点的记录

代码：

#include<cstdio>#include<algorithm>using namespace std;int w,h,n;const int maxn=100000+10;void update(int x,int a,int w,double& L,double& R){if(a==0){if(x<=0||x>=w)R=L-1;}else if(a>0){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);}}struct Event{double x;int type;bool operator < (const Event& a) const{return x<a.x||(x==a.x && type>a.type);} };Event event[2*maxn];int main(){int T,x,y,a,b;scanf("%d",&T);while(T--){scanf("%d%d%d",&w,&h,&n);int e=0;for(int i=0;i<n;i++){scanf("%d%d%d%d",&x,&y,&a,&b);double L=0,R=1e9;update(x,a,w,L,R);update(y,b,h,L,R);if(R>L){event[e++]=(Event){L,0};event[e++]=(Event){R,1};}}sort(event,event+e);int cnt=0,ans=0;for(int i=0;i<e;i++){if(event[i].type==0)ans=max(ans,++cnt);elsecnt--;//遇到左端就加一，遇到右端就减一 ；真的是好精彩的算法；太美妙了 }printf("%d\n",ans);}return 0;}