BOJ 394
Problem I. The Longest Sequence of
Rectangles
Description
A rectangle is specified by a pair of coordinates (x1 , y1) and (x2 , y2)
indicating its lower-left and upper-right corners (x1 <= x2 and y1 <= y2). For
a pair of rectangles, A = ((X_A1, Y_A1), (X_A2, Y_A2)) and B = ((X_B1, Y_B1),
(X_B2, Y_B2)), we define A <= B if X_A2<X_B1 and Y_A2<Y_B1. Given a
number of rectangles on the plane, you are asked to find the length L of the
longest sequence of rectangles (A1, A2, ..., AL) such that A1 <= A2 <= ...
<= AL.
Input
The input contains multiple test cases. The first line of the input is an integer
T, indicating the number of test cases. Each test case begins with a line
containing a single integer n (1 <= n <= 100000), indicating the number of
rectangles. Each of the next n lines contains four integers x1, y1, x2, y2 (-
1000000 <= x1 < x2 <= 1000000, -1000000 <= y1 < y2 <= 1000000),
indicating the lower left and upper right corners of a rectangle.
Output
For each input test case, output a single integer indicating the length of the
longest sequence.
Sample Input
2
3
1 5 2 8
3 -1 5 4
10 10 20 20
2
2 1 4 5
6 5 8 10
Sample Output
2
1
解题思路：（此题的原型就是上一篇日志nlogn最长上升子序列(LIS) ）显然，最终的答案序列中矩形的左下角和右上角点坐标一定是按x严格递增的。这就给了我们将二维问题降至一维的思路。将每个矩形视为两个点（左下角和右上 角）的组合。将这2N 个点按照x坐标排序。这样可以将问题从二维空间降至一维空间：维护数组y[m] ，表示长度为m矩形序列的最后一个矩形y坐标最小是多少。依次遍历这排序后的2N 个点。若当前点为矩形的左下角，二分查找该点的y坐标在F数组中的位置。若当前点为矩形的右上角，则用其左下角的值更新F 数组。之所以不可以立即更新F 数组，是因为要求矩形序列中后一个矩形的左下角严格在前一个的右上方。
