uva 1382 Distant Galaxy （枚举）

                          uva 1382 Distant Galaxy

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?

Input

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

N = 0

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

题目大意：给出n个点，问说一个平行与x轴和y轴的矩形，最多能有多少个点落在边上。

解题思路：

先枚举上下边界，然后从左到右扫，扫描一遍所有的点，计算L， on， on2数组，枚举右边界，维护on[i] - L[i]的最大值。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int n, y[1005], on[1005], on2[1005], L[1005];  
struct point {
	int x, y;
};  
int cmp(point a, point b) {
	return a.x < b.x;
}
point p[1005];
int getAns(){  
	int ans = 0;  
	sort(p, p + n, cmp);  
	sort(y, y + n);  
	int m = unique(y, y + n) - y;  //统计具有不同y坐标的点的个数
	if(m <= 2) return n;  
	for(int a = 0; a < m; a++)          
		for(int b = a + 1; b < m; b++){  
			int miny = y[a], maxy = y[b], k = 0;  //枚举上下边
			memset(on, 0, sizeof(on));  
			memset(on2, 0, sizeof(on2));  
			memset(L, 0, sizeof(L));  
			for(int i = 0; i < n; i++){         
				if(!i || p[i].x != p[i-1].x){  
					k++;  
					if(k > 1) L[k] = L[k - 1] + on2[k - 1] - on[k - 1]; //on2记录边界,on没记录边界，on2-on就是边界上的点的个数， 加上上一条竖线左侧，就是当前竖线左侧点的个数 
				}  
				if(p[i].y < maxy && p[i].y > miny) on[k]++;  
				if(p[i].y <= maxy && p[i].y >= miny) on2[k]++;  
			}  
			if(k <= 2) return n;  
			int Max = 0;  
			for(int i = 1; i <= k; i++){   
				ans = max(ans, L[i] + on2[i] + Max);  //维护Max
				Max = max(Max, on[i] - L[i]);  
			}  
		}  
	return ans;  
}  
int main(){  
	int Case = 1;  
	while(scanf("%d", &n), n){  
		for(int i = 0; i < n; i++){  
			scanf("%d %d", &p[i].x, &p[i].y);  
			y[i] = p[i].y;  
		}  
		printf("Case %d: %d\n", Case++, getAns());  
	}  
	return 0;  
}