Poj 1458 common subsequence

Common subsequence

Time limit:1000 ms		Memory limit:10000 K
Total submissions:39058		Accepted:15739

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         abfcabprogramming    contest abcd           mnp

Sample output

420

Question: Find the longest common subsequence;

Solution:

Longest Common subsequence (LCS ):

String: S: s1s2s3s4 ...... Si, T: t1t2t3t4 ...... TJ;

Define an array DP [I] [J] to indicate s1s2s3s4 ....... si and t1t2t3t4 .......... the longest common subsequence corresponding to TJ;

Then, the longest common subsequence corresponding to s1s2s3s4 ...... Si + 1 and t1t2t3t4 ...... TJ + 1 may be:

(1) If Si + 1 = TJ + 1, s1s2s3s4 ....... si and t1t2t3t4 .......... the longest common subsequence corresponding to TJ + 1;

(2) s1s2s3s4 ...... Si + 1 and t1t2t3t4 ...... TJ correspond to the longest common subsequence;

(3) s1s2s3s4 ...... Si and t1t2t3t4 ...... The Longest Common subsequence corresponding to TJ + 1;

So the state transition equation of DP is DP [I] [J] = max (DP [I-1] [J-1] + (A [I] = B [J]), max (DP [I-1] [J], DP [I] [J-1])

#include <iostream>#include <string.h>using namespace std;char a[1000],b[1000];int dp[1000][1000];int max(int a,int b){return a>b?a:b;}int LCS(char *a,char *b){int n=strlen(a),m=strlen(b);for (int i=0;i<n;i++)dp[i][0]=0;for (int j=0;j<m;j++)dp[0][j]=0;for (int i=0;i<n;i++)for (int j=0;j<m;j++){if (a[i]==b[j])dp[i+1][j+1]=dp[i][j]+1;elsedp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);}return dp[n][m];}int main(){while (cin>>a>>b){cout<<LCS(a,b)<<endl;}return 0;} 

Poj 1458 common subsequence