Shanghai 2006 / UVa 1382 Distant Galaxy (枚举&扫描&动态维护)

1382 - Distant Galaxy

Time limit: 3.000 seconds 

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=460&page=show_problem&problem=4128

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 integerN , (1N100) , the number of star systems on the telescope. N lines follow, each line consists of two integers: theX and Y coordinates of theK -th planet system. The absolute value of any coordinate is no more than10⁹ , and you can assume that the planets are arbitrarily distributed in the universe.

N = 0 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

思路：

枚举矩形的上下界，然后选择左右边界。 对于确定的左边界left和右边界right， 假设是下图的R3是left,  L3是right，那么数量为：

L1 + L2 + L3 - (R1+R2) + R3.  

为了要使得以L3为右边界的矩形上的点最多，那么应该使得 R3-(R1+R2)的值最大。

所以，枚举上下边界，然后枚举右边界j， 同时维护保存j左边的R3-(R1+R2)的最大值，O(n^3)确定答案。

完整代码：

/*0.025s*/

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 105;

struct Point
{
	int x, y;
	bool operator < (const Point& rhs) const
	{
		return x < rhs.x;
	}
} P[maxn];
int n, m, y[maxn], on[maxn], on2[maxn], left[maxn];

int solve()
{
	sort(P, P + n);
	sort(y, y + n);
	m = unique(y, y + n) - y; // 所有不同的y坐标的个数
	if (m <= 2) return n; // 特判，最多两种不同的y
	int ans = 0;
	for (int a = 0; a < m; a++)
		for (int b = a + 1; b < m; b++)
		{
			int ymin = y[a], ymax = y[b]; // 计算上下边界分别为ymin和ymax时的解
			// 计算left, on, on2
			int k = 0;
			for (int i = 0; i < n; i++)
			{
				if (i == 0 || P[i].x != P[i - 1].x) // 一条新的竖线
				{
					k++;
					on[k] = on2[k] = 0;
					left[k] = (k == 0 ? 0 : left[k - 1] + on2[k - 1] - on[k - 1]);
				}
				if (P[i].y > ymin && P[i].y < ymax) on[k]++;
				if (P[i].y >= ymin && P[i].y <= ymax) on2[k]++;
			}
			if (k <= 2) return n; // 特判，最多两种不同的x
			int M = 0;
			for (int j = 1; j <= k; j++)
			{
				ans = max(ans, left[j] + on2[j] + M);
				M = max(M, on[j] - left[j]);//贪心嘛，动态维护on[j] - left[j]
			}
		}
	return ans;
}

int main()
{
	int kase = 0;
	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", ++kase, solve());
	}
	return 0;
}