Poj 3264 balanced lineup (ST algorithm entry)

Reprinted please indicate the source, thank youHttp://blog.csdn.net/acm_cxlove/article/details/7854526
By --- cxlove

Question: maximum and minimum values of the query Interval

Http://poj.org/problem? Id = 3264

In the past, only the line segment tree approach, the establishment of nlgn, and the query of lgn were used.

Rmq, as a common question, must be learned and can be used as a tool.

StAlgorithmIs another efficient algorithm for solving the maximum value of a range. The processing of nlgn can achieve O (1) queries, and the constant of the Line Segment tree is also very large.

It is a dynamic planning method.
Take the minimum value as an example. A is the search array.
Record the minimum value in the interval [I, I + 2 ^ J-1] (lasting 2 ^ J) using a two-dimensional array f (I, j. F [I, 0] = A [I];
So for any group (I, j), F (I, j) = min {f (I, J-1), F (I + 2 ^ (J-1 ), j-1)} to use dynamic planning calculations.
This algorithm is clever not in the establishment of dynamic planning, but in its query: its query efficiency is O (1 ).
Suppose we require the minimum value of A in the range [M, N] and find a number k to make 2 ^ k <n-m + 1.
In this way, we can divide this interval into two parts: [M, m + 2 ^ k-1] and [N-2 ^ k + 1, N]. we found that these two intervals have been initialized.
The preceding range is F (M, K), and the following interval is F (n-2 ^ k + 1, K ).
In this way, you can know the minimum value of the entire interval by looking at the minimum values of the two intervals!

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        # define n100005 using namespace STD; int N, Q, A [n]; int MX [N] [18], Mn [N] [18]; void rmq_init () {int M = floor (log (double) N)/log (2.0); For (INT I = 1; I <= N; I ++) MX [I] [0] = Mn [I] [0] = A [I]; for (INT I = 1; I <= m; I ++) for (Int J = N; j --) {MX [J] [I] = Mx [J] [I-1]; mn [J] [I] = Mn [J] [I-1]; If (J + (1 <(I-1) <= N) {MX [J] [I] = max (MX [J] [I], MX [J + (1 <(I-1)] [I-1]); mn [J] [I] = min (Mn [J] [I], Mn [J + (1 <(I-1)] [I-1]) ;}} int rmq_query (int l, int R) {int M = floor (log (double) (R-l + 1)/log (2.0 )); int max = max (MX [l] [m], MX [R-(1