LA 3695 Distant Galaxy

 

3695 - Distant Galaxy

Time limit: 3.000 seconds

You are observing a distant galaxy using a telescope above the Astronomy Tower, and you think that a rectangle drawn in that galaxy whose edges are parallel to coordinate axes and contain maximum star systems on its edges has a great deal to do with the mysteries of universe. However you do not have the laptop with you, thus you have written the coordinates of all star systems down on a piece of paper and decide to work out the result later. Can you finish this task?

 <tex2html_verbatim_mark>

Input 

There are multiple test cases in the input file. Each test case starts with one integer N <tex2html_verbatim_mark>, (1N100) <tex2html_verbatim_mark>, the number of star systems on the telescope. N <tex2html_verbatim_mark>lines follow, each line consists of two integers: the X <tex2html_verbatim_mark>and Y <tex2html_verbatim_mark>coordinates of the K <tex2html_verbatim_mark>-th planet system. The absolute value of any coordinate is no more than 10⁹ <tex2html_verbatim_mark>, and you can assume that the planets are arbitrarily distributed in the universe.

N = 0 <tex2html_verbatim_mark>indicates the end of input file and should not be processed by your program.

Output 

For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.

Sample Input 

10 
2 3 
9 2 
7 4 
3 4 
5 7 
1 5 
10 4 
10 6 
11 4 
4 6 
0

Sample Output 

Case 1: 7

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define mod 1000000007
#define ll long long
#define maxn 105
#define pi acos(-1.0)
//const int maxk = 10 + 1, inf = ~0U >> 2;
int n, m, ans,s;
struct point{
    int x, y;
}a[maxn];
bool cmp(point a, point b){ return a.x < b.x; }
int px[maxn], p[maxn],lf[maxn],on[maxn],on2[maxn];
int main(){
    int cas = 1;
    while (~scanf("%d",&n)&&n){
        int t = 0, tt = 0;
        for (int i = 0; i < n; i++){
            scanf("%d%d", &a[i].x, &a[i].y);
            px[tt++] = a[i].y;
        }
        sort(a, a + n, cmp);
        sort(px, px + tt);
        if(tt)p[t++] = px[0];
        for (int i = 1; i < tt; i++)if(px[i]!=px[i-1])p[t++]=px[i];
        printf("Case %d: ", cas++);
        if (t <= 2){ printf("%d\n", n); continue; }    
        ans = 0;
        for (int i = 0; i < t;i++)
        for (int j = i+1; j < t; j++){
            //memset(lf, 0, sizeof lf);
            int h = 0;
            for (int k = 0; k < n; k++){
                if (k==0||a[k].x!=a[k-1].x){
                    h++;
                    on[h] = on2[h] = 0;
                    lf[h] = lf[h - 1] + on2[h - 1] - on[h - 1];
                }
                if (a[k].y >p[i] && a[k].y < p[j])on[h]++;
                if (a[k].y >= p[i] && a[k].y <= p[j])on2[h]++;
            }
            if (h <= 2){ ans = n; i = t; break; }
            int res = 0;
            for (int k = 1; k <= h; k++){
                ans = max(ans, on2[k] + res + lf[k]);
                res = max(res, on[k] - lf[k]);
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}