Common Subsequence HDU dp

Common Subsequence Time Limit: 2000/1000 ms (Java/Other) Memory Limit: 65536/32768 K (Java/Other) Total Submission (s): 11 Accepted Submission (s ): 6 Problem Description A 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 incres Asing 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 [cpp] <SPAN style = "FONT-SIZE: 14px" >#include <cstdio> # Include <cstring> using namespace std; int dp [1001] [1001]; int main () {char pre [1001]; char pos [1001]; while (scanf ("% s", pre, pos )! = EOF) {memset (dp, 0, sizeof (dp); int l1 = strlen (pre); int l2 = strlen (pos); for (int I = 0; I <l1; I ++) {for (int j = 0; j <l2; j ++) {if (pre [I] = pos [j]) dp [I + 1] [j + 1] = dp [I] [j] + 1; else if (dp [I + 1] [j]> dp [I] [j + 1]) dp [I + 1] [j + 1] = dp [I + 1] [j]; else dp [I + 1] [j + 1] = dp [I] [j + 1] ;}} printf ("% d \ n ", dp [l1] [l2]);} return 0 ;}</SPAN >#include <cstdio >#include <cstring> using namespace std; int dp [1001] [1001]; int main () {char pre [1001]; char pos [1001]; while (scanf ("% s", pre, pos )! = EOF) {memset (dp, 0, sizeof (dp); int l1 = strlen (pre); int l2 = strlen (pos); for (int I = 0; I <l1; I ++) {for (int j = 0; j <l2; j ++) {if (pre [I] = pos [j]) dp [I + 1] [j + 1] = dp [I] [j] + 1; else if (dp [I + 1] [j]> dp [I] [j + 1]) dp [I + 1] [j + 1] = dp [I + 1] [j]; elsewww.2cto.com dp [I + 1] [j + 1] = dp [I] [j + 1] ;}} printf ("% d \ n ", dp [l1] [l2]);} return 0 ;}