Ultraviolet A 10635 prince and princess (LIS)

It is also a classic question in the training guide ~~

Http://uva.onlinejudge.org/external/106/10635.html

There are two sequences with the length of p + 1 and q + 1, each of which has different elements and is 1 ~ An integer of N ^ 2. Calculate the LCS of the two sequences.

Analysis: The O (PQ) Complexity of LCS is obviously too slow. Note that "each sequence has different elements and is 1 ~ N ^ 2 integer ", so there is a clever conversion, re-element in

By 1 ~ P + 1 number, and B is also the same ing number. Because a is an incremental sequence, LCS is the longest incremental subsequence of B, that is, Lis. in this way, the LIS problem of B can be converted.

Solved in O (nlogn) time.

LIS:

Set DP [I] to the length of the longest ascending subsequence ending with a [I.

O (N ^ 2) solution: DP [I] = max {0, DP [J] | j <I, AJ <AI} + 1;

O (nlogn) solution: Assume that two States A and B have met AA <AB and DP [a] = DP [B]; in this case, I select a in all subsequent states, which is not inferior to B. In this way, for the same Dp value, you only need to keep the smallest one of A (seeCode).

View code

 1   /*  2  Author: zhaofa Fang  3   Lang: C ++  4   */  5 # Include <cstdio> 6 # Include <cstdlib> 7 # Include <sstream> 8 # Include <iostream> 9 # Include <cmath> 10 # Include <cstring> 11 # Include <algorithm> 12 # Include < String > 13 # Include <utility> 14 # Include <vector> 15 # Include <queue> 16 # Include <stack> 17 # Include <map> 18 # Include < Set > 19   Using   Namespace  STD;  20  21 Typedef Long   Long  Ll;  22   # Define Debug (x) cout <# x <':' <x <Endl 23   # Define Rep (I, n) for (INT I = 0; I <(n); I ++) 24   # Define For (I, S, T) for (INT I = (s); I <= (t); I ++) 25   # Define PII pair <int, int> 26  # Define PB push_back 27   # Define MP make_pair 28   # Define Fi first 29   # Define Se second 30   # Define Lowbit (x) (X & (-x )) 31   # Define INF (1 <30) 32   33   Const  Int Maxn = 250 * 250 + 10  ;  34   Int  A [maxn], B [maxn], POS [maxn], DP [maxn];  35   36   Int Lis ( Int  N)  37   {  38      For ( Int I = 1 ; I <= N; I ++) A [I] = inf, DP [I] = 0  ;  39       For ( Int I = 0 ; I <n; I ++ )  40   {  41           Int K = lower_bound (a + 1 , A +1 + N, B [I])- A;  42 A [k] = B [I];  43 DP [I] = K;  44   }  45       Int MX =- 1  ;  46 Rep (I, n) MX = Max (DP [I], MX );  47      Return  MX;  48   }  49   Int  Main ()  50   {  51       //  Freopen ("in", "r", stdin );  52       Int  T;  53 Scanf ( " % D  " ,& T );  54   Rep (CAS, T)  55   {  56 Printf ( "  Case % d:  " , CAS + 1  );  57           Int  N, P, Q;  58 Scanf ( "  % D  " , & N, & P ,& Q );  59 Memset (Pos, 0 , Sizeof  (POS ));  60 Rep (I, P + 1  )  61   {  62 Scanf ( " % D  " ,& A [I]);  63 Pos [A [I] = I + 1  ;  64   }  65 Rep (I, q + 1  )  66   {  67 Scanf ( "  % D " ,& B [I]);  68 B [I] = Pos [B [I];  69   }  70 Printf ( "  % D \ n  " , Lis (q + 1  ));  71   }  72       Return  0  ;  73 }