Common Subsequence --- Longest Common Subsequence and longest Common Subsequence

Common Subsequence --- Longest Common Subsequence and longest Common Subsequence
Problem DescriptionA subsequence of a given sequence is the given sequence with some elements (possible none) left out. given a sequence X = <x1, x2 ,..., xm> another sequence Z = <z1, z2 ,..., zk> is a subsequence of X if there exists a strictly increasing sequence <i1, i2 ,..., ik> of indices of X such that for all j = 1, 2 ,..., k, xij = zj. for example, Z = <a, B, f, c> is a subsequence of X = <a, B, c, f, B, c> with index sequence <1, 2, 4, 6>. given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.
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.

 

Sample InputAbcfbc abfcabprogramming contest abcd mnp

 

Sample Output420 the code is as follows:

# Include <stdio. h> # include <string. h ># define N 2000int max (int a, int B) {if (a> B) return a; else return B;} char a [N], B [N]; int ch [N] [N] = {0}; int main () {int I, j, m, n; while (scanf ("% s",, B )! = EOF) {memset (ch, 0, sizeof (ch); // obtain zero; m = strlen (a); n = strlen (B ); for (I = 1; I <= m; I ++) {for (j = 1; j <= n; j ++) {if (a [I-1] = B [J-1]) ch [I] [j] = ch [I-1] [J-1] + 1; // The reason is shown in the table below; else ch [I] [j] = max (ch [I-1] [j], ch [I] [J-1]);} printf ("% d \ n ", ch [m] [n]);} return 0 ;}

Example

A = abcfbc

B = abfcab

Returns the longest common subsequence.

A \ B	A0 =	A1 = B	A2 = c	A3 = f	A4 = B	A5 = c	A6	A7	A8
B0 =	1	1	1	1	1	1	For the method to evaluate the space value, see the code. If (a [I-1] = B [J-1]) Ch [I] [j] = ch [I-1] [J-1] + 1; Else Ch [I] [j] = max (ch [I-1] [j], ch [I] [J-1]);
B1 = B	1	2	2	2	2	2
B2 = f	1	2	2	3	3	3
B3 = c	1	2	3	3	3	4
B4 =	1	2	3	3	3	4
B5 = B	1	2	3	3	4	4
B6
B7
B8