Problem B (subsequence problem)

Problem descriptiona subsequence of a given sequence is the given sequence with some elements (possible none) left out. given a sequence X = & lt; x1, x2 ,..., XM & gt; another sequence z = & lt; Z1, Z2 ,..., ZK & gt; Is a subsequence of X if there exists a strictly increasing sequence & lt; I1, I2 ,..., ik & gt; of indices of X such that for all j = 1, 2 ,..., k, Xij = ZJ. for example, Z = & lt; A, B, F, C & gt; Is a subsequence of X = & lt; A, B, C, F, B, C & gt; with index sequence & lt; 1, 2, 4, 6 & gt ;. given two sequences x and y the problem is to find the length of the maximum-length common subsequence of X and Y. <br> the program input is from a text file. each data set in the file contains two strings representing the given sequences. the sequences are separated by any number of white spaces. the input data are correct. for each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line. <br>

 

Sample inputabcfbc abfcabprogramming contest abcd mnp sample output420 question: two strings are given to identify the maximum number of identical characters in the two strings. Idea: Set the string a [0... n], B [0... m], string a corresponds to the row of the Two-dimensional array num, string B Corresponds to the column of the Two-dimensional array num. Then the status code after each record comparison:

# Include <iostream>
# Include <cstring>
Using namespace STD;

Char str1 [1005];
Char str2 [1005];
Int DP [1005] [1005];

Int main ()
{
Int I, J;
Int len1, len2;
While (CIN> str1> str2)
{
Len1 = strlen (str1 );
Len2 = strlen (str2 );
For (I = 0; I <len1; I ++)
{
DP [I] [0] = 0;
}
For (I = 0; I <len2; I ++)
{
DP [0] [I] = 0;
}
For (I = 1; I <= len1; I ++)
{
For (j = 1; j <= len2; j ++)
{
If (str1 [I-1] = str2 [J-1])
{
DP [I] [J] = DP [I-1] [J-1] + 1;
}
Else
{
DP [I] [J] = max (DP [I-1] [J], DP [I] [J-1]);
}
}
}
Cout <DP [len1] [len2] <Endl;
}
Return 0;
}

There are multiple ways to do this.

 

Problem B (subsequence problem)