Algorithm for the longest common subsequence for small character sets

Generally, for two strings, the lengths are N and M, and their time complexity is O (nm).

However, for small character sets, the complexity can be reduced to O (n^2+km), where n is a shorter length of two strings. This method is much better than O (nm) for a large difference in the length of two strings.

It is assumed that all characters are lowercase letters, so that they conform to the premise of a small character set. Set a shorter string to S1, and a longer string of S2. The string subscript starts at 1.

S2 string The first character to the right of each position can be preprocessed by O (km). where k is the number of characters in a small character set, and M is the length of the longer string.

Use Next[i][j] to represent the position of the first (char) (' a ' +j) on the right of S2[i].

Set Dp[i][j] means that S1 matches the first I bit, and the longest common subsequence of length J matches S2 to the leftmost position. Length (S2) +1 if it does not exist.

Dp[i][0] = 0

If S2 's dp[i-1][j-1] has the same position as the first to s1[i] on the right, then dp[i][j] = min{dp[i-1][j], Next[dp[i-1][j-1]][s1[i]]}.

otherwise dp[i][j] = Dp[i-1][j].

For each DP that is not-1, record J, and finally take a maximum of the longest common subsequence.

Therefore, the total complexity is O (n^2+km).

Specific implementation:

#include <cstdio>#include <cstring>#include <algorithm>using namespaceStd;const int inf=0x3f3f3f3f; Const INT MAXN=1005; Const INT MAXM=1000005; charS1[maxn],s2[maxm];intDp[maxn][maxn];int next[maxm][26]; IntMain () {scanf ("%s%s", s1+1,s2+1), int l1=strlen (s1+1); int L2=strlen (s2+1); for (int i=0;i<maxm;i++) for (int j=0;j<26;j++) next[i][j]=l2+1, for (int. i=0;i<maxn;i++) for (int j=0;j<maxn;j++) dp[i] [J]=l2+1; for (int i=l2-1;i>=0;i--) {for (int j=0;j<26;j++) {char cc= ' a ' +J; if (S2[I+1]==CC) NEX T[i][j]=i+1, Else next[i][j]=next[i+1][j];} for (int i=1;i<=l1;i++) dp[i][0]=0, int ans=0; for ( int i=1;i<=l1;i++) {for (int j=1;j<=i;j++) {if (next[dp[i-1][j-1]][s1[i]-' a ']!=l2+1) dp[i][j]=min (dp[ i-1][j],next[dp[i-1][j-1]][s1[i]-' a ']); else dp[i][j]=dp[i-1][j]; if (dp[i][j]!=l2+1) ans=max (ANS,J);}} printf ("%d\n", ans); return 0;}  

Algorithm for the longest common subsequence for small character sets