Pku1458 longest Public String DP

Common subsequence

Time limit: 1000 ms		Memory limit: 10000 K
Total submissions: 20435		Accepted: 7752

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 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.

Input

The program input is from the STD input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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 Input

Abcfbc
Abfcab

Programming
Contest

ABCD
MNP

Sample output

4

2

0

 

 

 

DP

Classic applications
:

Longest Public String
.

Transfer Equation
:
(RES [I] [J]:

Indicates
Stringa

Before
I

Characters and
Stringb

Before
J

Maximum length of Public strings
)

Res [I] [J] = 0


(I = 0 | j = 0)

Res [I] [J] = res [I-1] [J-1] + 1


(Stringa [I] = stringb [J])

Res [I] [J] = max (RES [I] [J-1], Res [I-1] [J])
(Stringa [I]! = Stringb [J])

 

The Code is as follows:
:

 

# Include <stdio. h> <br/> # include <string. h> <br/> # define maxn 1005 <br/> int main () <br/> {<br/> int I, j, La, LB, res [maxn] [maxn]; char a [maxn], B [maxn]; <br/> while (scanf ("% S % s", a, B )! = EOF) <br/>{< br/> memset (Res, 0, sizeof (RES); <br/> La = strlen (); lb = strlen (B); <br/> for (I = 0; I <la; ++ I) <br/> for (j = 0; j <Lb; ++ J) <br/> if (a [I] = B [J]) res [I + 1] [J + 1] = res [I] [J] + 1; <br/> else res [I + 1] [J + 1] = res [I] [J + 1]> res [I + 1] [J]? Res [I] [J + 1]: res [I + 1] [J]; <br/> printf ("% d/N ", res [la] [LB]); <br/>}< br/> return 0; <br/>}< br/>