Ultraviolet A 10066-the Twin Towers

The longest common subsequence is required.

Question type: DP/LCs

Question Analysis:

Didn't I wake up? Why is it clearly written: // print a blank line after the output of each data set, instead of a blank line between each group of data. It turned out to be a long time.

 Code:

① The version of the memory-based search.

// In fact, the topic is about the longest common subsequence. Fuck. <Br/> # include <cstdio> <br/> # include <cstring> <br/> using namespace STD; <br/> int N1, N2; <br/> # define maxn 102 <br/> int d [maxn] [maxn]; <br/> int vis [maxn] [maxn]; <br/> int A [maxn], B [maxn]; <br/> int dp (int I, Int J) <br/>{< br/> if (I <0 | j <0) <br/>{< br/> return 0; <br/>}< br/> If (vis [I] [J]) return d [I] [J]; <br/> vis [I] [J] = 1; <br/> int Max; <br/> if (a [I] = B [J]) <br/> max = dp (I-1, J-1) + 1; <br/> else <Br/> {<br/> max = dp (I, J-1)> dp (I-1, j )? DP (I, J-1): dp (I-1, J); <br/>}< br/> return d [I] [J] = max; <br/>}< br/> inline int RINT () {int X; scanf ("% d", & X); Return X ;} <br/> int main () <br/> {<br/> int T = 1; <br/> while (1) // when at least one is not 0 <br/>{< br/> n1 = RINT (); N2 = RINT (); <br/> If (! N1 &&! N2) break; <br/> // If (T! = 1) printf ("/N"); <br/> memset (VIS, 0, sizeof (VIS); <br/> for (INT I = 0; I <N1; I ++) <br/> A [I] = RINT (); <br/> for (INT I = 0; I <N2; I ++) <br/> B [I] = RINT (); <br/> int ans = dp (n1-1, n2-1 ); <br/> printf ("Twin Towers # % d/N", t ++); <br/> printf ("Number of tiles: % d/N ", ans); <br/> printf ("/N"); // print a blank line after the output of each data set. <br/>}< br/>}

②

From zcoder, http://www.zhaocong89.com /? P = 498

Returns the version of the state transition equation.

# Include <iostream> <br/> # include <fstream> <br/> # include <cstring> <br/> # include <cstdlib> <br/> using namespace STD; <br/> # ifndef online_judge <br/> ifstream fin ("test. in "); <br/> ofstream fout (" test. out "); <br/> # else <br/> # define fin CIN <br/> # define fout cout <br/> # endif <br/> const int maxn = 105; <br/> int n, m; <br/> int T1 [maxn]; <br/> int T2 [maxn]; <br/> int d [maxn] [maxn]; <br/> int main () <br/> {<br/> I NT ncase = 0; <br/> while (Fin> N> M &&(! (N = 0 & M = 0) <br/>{< br/> // If (ncase) <br/> // fout <Endl; <br/> memset (D, 0, sizeof (d); <br/> for (INT I = 1; I <= N; I ++) <br/> fin> T1 [I]; <br/> for (INT I = 1; I <= m; I ++) <br/> fin> T2 [I]; <br/> for (INT I = 1; I <= N; I ++) <br/> {<br/> for (Int J = 1; j <= m; j ++) <br/> {<br/> If (T1 [I] = t2 [J]) <br/> d [I] [J] = d [I-1] [J-1] + 1; <br/> else <br/> d [I] [J] = max (d [I-1] [J], d [I] [J-1]); <br/>}< br/> fout <"Twin Towers #" <(++ ncase) <Endl; <br/> fout <"Number of tiles:" <D [N] [m] <Endl; <br/>}< br/> return 0; <br/>}