Hdu4521 Xiaoming sequence (line segment tree + dp)

Question: The meaning of the question is very simple. Given a sequence, find the length of the maximum incrementing subsequence, and the difference between the two adjacent subsequences must be greater than D.

Analysis: just to get the question, just add a restriction condition to the longest incrementing subsequence and follow the idea of the longest incrementing subsequence, then DP [I] = max (DP [I], DP [J] + 1), (I-j> D & A [I]> A [J])

From the state transition equation, we can see that the key to the problem is that the maximum Dp value of the maintenance interval (0, I-d-1) is smaller than a [I].

We use the value of a [I] As the subscript to guarantee the current range (0, a [I]-1) if the Dp value in all satisfies (I-j> D), then DP [I] is equal to the maximum value of the range (0, a [I]-1, this can be solved using the line segment tree.

Hdu4521

# Include <iostream> # Include <Algorithm> # Include <Stdio. h> # Include < String . H> # Include <Stdlib. h>Using   Namespace  STD;  Const   Int N = 100000 + 10  ;  Struct  Node {  Int  L, R, Max;} p [n * 3  ];  Int  A [n], d, n; Int  DP [N];  Void Build ( Int S, Int T, Int  K) {P [K]. L = S; P [K]. r = T; P [K]. Max = 0  ;  If (S = T) Return  ;  Int Mid = (S + T)> 1 , KL = k < 1 , Kr = KL + 1  ; Build (S, mid, KL); Build (Mid + 1  , T, KR );}  Void Insert ( Int K, Int S, Int  Val ){  If (P [K]. L = P [K]. R & P [K]. L = S) {P [K]. Max = Max (P [K]. Max, Val );  Return  ;}  Int Mid = (P [K]. L + P [K]. R)> 1 , KL = k < 1 , Kr = KL + 1  ;  If (S <= Mid) insert (KL, S, Val );  Else  Insert (KR, S, Val); P [K]. Max = Max (P [Kl]. Max, P [Kr]. max );} Int Query ( Int K, Int S, Int  T ){  If (S <= P [K]. L & T> = P [K]. R ){  Return  P [K]. Max ;}  Int Mid = (P [K]. L + P [K]. R)> 1 , KL = k < 1 , Kr = KL + 1 ;  Int A = 0 , B = 0  ;  If (S <= mid) A = Query (KL, S, T );  If (T> mid) B = Query (KR, S, T );  Return  Max (a, B );}  Int  Main (){  While (Scanf ( " % D  " , & N, & D) = 2  ){  Int Maxa = 0  ;  For ( Int I = 1 ; I <= N; ++ I) {scanf (  "  % D  " , A + I ); ++ A [I]; maxa = Max (maxa, a [I]);} memset (DP,  0 , Sizeof  (DP); Build (  0 , Maxa, 1  );  Int Ans = 1  ;  For ( Int I = 1 ; I <= N; ++I ){  Int TMP = query ( 1 , 0 , A [I]- 1  ); DP [I] = Max (DP [I], TMP + 1  ); Ans = Max (ANS, DP [I]);  If (I-d> 0 ) Insert ( 1 , A [I-d], DP [I-d]); // Insert a Dp value that meets the subscript in the online segment tree  } Printf (  "  % D \ n  "  , ANS );}  Return   0  ;}