"Algorithm learning note" 72.LCS Max-father-in-law sub-sequence dynamic programming SJTU OJ 1065 Small M biological experiment 1

A very simple DP

If DP[I,J] represents the similarity between the two segments from 0 to I and from 0 to J,

Then you know that each dp[i,j] is transformed by three states.

The first kind of dna1[i]==dna2[j]

DP[I-1,J-1] + 1 length plus 1

The second, or else from the bottom two states.

DP[I][J-1] and dp[i-1][j]//note because it's sequential traversal, both of which have been computed

Take the two to the maximum.

#include <iostream>#include<cstring>#include<algorithm>using namespacestd;//Longest common sub-sequence length LCS dpChardna1[ ++Ten];Chardna2[ ++Ten];intlen1, Len2;intdp[ ++Ten][ ++Ten];//represents the similarity of dna1 from 1th to I and dna2 from first position to Jvoidinit () {cin>>Dna1; CIN>>Dna2; Len1=strlen (DNA1); Len2=strlen (DNA2);}intbuild () {memset (DP,0,sizeof(DP));  for(inti =1; I <= len1; ++i) {         for(intj=1; J <= Len2; ++j) {            BOOLOK = (dna1[i-1]==dna2[j-1]); DP[I][J]=Max (Dp[i-1][j-1] +OK, max (Dp[i-1][j], dp[i][j-1]                )                ); }    }    returndp[len1][len2];}intMainintargcChar Const*argv[])    {init (); cout<<build () <<Endl; return 0;}

"Algorithm learning note" 72.LCS Max-father-in-law sub-sequence dynamic programming SJTU OJ 1065 Small M biological experiment 1