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

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本文探讨了如何通过算法找出望远镜观测到的遥远星系中，包含最多星系的矩形边界问题。该算法首先对星系坐标进行排序，并通过枚举矩形的上下边界来确定可能的最大星系数量。

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?

$\epsfbox{p3694.eps}$

Input

There are multiple test cases in the input file. Each test case starts with one integerN , (1 $\le$ N $\le$ 100) , 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

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;
}