UVA 1398 - Meteor

 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 meteor m_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 of m_i . The point p_i = (x_i, y_i) is represented by X -coordinate x_i and Y -coordinate y_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 meteor m_i will be at the position (0, 5.5) at time t = 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 , and p₅ 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 integers w 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 integers x_i, y_i, a_i , and b_i ; (x_i, y_i) is the starting point p_i and (a_i, b_i) is the nonzero velocity vector v_i of the i -th meteor; x_i and y_i are integer values between -200,000 and 200,000, and a_i and b_i are integer values between -10 and 10. Note that at least one of a_i and b_i is not zero. These four values are separated by single spaces. We assume that all starting points p_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

有n个流星，各自以不同速度朝不同方向运动，求所有时刻，同时在矩形区域中的最大流星数，边界上的不算。

所有流星，能求出进入矩形和离开矩形区域的时间。这个时间段可以在数轴上表示出来，只要求出数轴上的点被不同时间段重叠次数最大的值即可。

从左往右扫描，遇到时间起点计数++,遇到时间终点计数--，维护此过程中的最大计数。

注意这里有一个问题，对于终点和起点重合的时间点t，先--后++,因为t时刻计算完的计数值，实际上是刚过t的时刻在矩阵中的流星数。终点的应当减去，起点的加上，如果先加后减，则把将要离开的流星也算上了，会导致max值更新，导致错误答案。

对于除法，若只需计算后的相对大小关系，为提高精度避免浮点运算，可以将被除数乘以所有除数的最小公倍数（应在整数范围内）。

 

#include <cstdlib>
#include <iostream>
#include <stdio.h>
#include <algorithm>

using namespace std;

const int MAX = 100000 + 5;
const int MAX_T = 1000000000 + 5;

struct point
{
    int t,type;//t = -1 表示是下界 1表示上界 
    
    bool operator <(const point& p)const
    {
        return t < p.t || (t == p.t && type > p.type);//某一时间点在边界上的 不算 x实际 
    }
}p[MAX * 2];
 

int x,y,a,b,min_t,max_t; 

void caculate_t(int edg, int pos,int v)
{
    if(v == 0)
    {
        if(pos <= 0 || pos >= edg)max_t = -1;//注意 边界不能算 
        return;
    } 
    
    int t1 = max(0, -2520 / v * pos);
    int t2 = max(0, 2520 / v * (edg - pos));
    //printf("pos :%d  edg:%d v:%d t1:%d  t2:%d\n",pos,edg,v,t1,t2);
    min_t = max(min_t, min(t1, t2));
    max_t = min(max_t, max(t1, t2)); 
    
    //printf("pos :%d  edg:%d v:%d l:%d  r:%d======\n",pos,edg,v,min_t,max_t);
}

void get_result(int n)
{
     int m = 0,cnt = 0;
     for(int i = 0; i < n; i++)
     {
         if(p[i].type < 0)m = max(++cnt, m);
         else cnt--;
         
        
     }
     printf("%d\n",m);
}

int main(int argc, char *argv[])
{
    
    int t,w,h,n;
    
    
    scanf("%d",&t);
    
    while(t--)
    {
        scanf("%d%d%d",&w,&h,&n);
        int cnt = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d%d%d%d",&x,&y,&a,&b);
        
            min_t = 0; max_t = MAX_T;
            caculate_t(w, x, a);
            caculate_t(h, y, b);
        
            if(max_t > 0 && max_t > min_t)
            {
                p[cnt++] = (point){min_t, -1};
                p[cnt++] = (point){max_t, 1};
            }
        }
        
        sort(p, p + cnt);
        get_result(cnt);
        
        
        
    }
    
    //system("PAUSE");
    return EXIT_SUCCESS;
}