Poj1458--dp,lcs

Poj1458--dp,lcscommon subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 40529		Accepted: 16351

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 the Re exists a strictly increasing sequence < I1, I2, ..., ik > of indices of X such so for all J =,..., K, Xij = Zj. For example, Z = < A, b, F, C > are a subsequence of X = < A, b, C, F, B, C > with index sequence < 1, 2, 4, 6;. Given 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 was from the STD input. Each data set in the input contains the strings representing the given sequences. The sequences is separated by any number of white spaces. The input data is correct.

Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from th e beginning of a separate line.

Sample Input

ABCFBC         abfcabprogramming    contest ABCD           MNP

Sample Output

420
Test instructions: Ask for LCS, water problem!

#include <iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespacestd;Const intmaxn=1000100;CharS[MAXN],T[MAXN];intdp[2][MAXN];intMain () { while(SCANF ("%s%s", s,t)! =EOF) {        intLs=strlen (s), lt=strlen (t); Memset (DP,0,sizeof(DP)); Char*ss=s-1, *tt=t-1; Memset (DP,0,sizeof(DP));  for(intI=1; i<=ls;i++){             for(intj=1; j<=lt;j++){                if(Ss[i]==tt[j]) dp[i%2][j]=dp[(i+1)%2][j-1]+1; Elsedp[i%2][j]=max (dp[(i+1)%2][j],dp[i%2][j-1]); }} printf ("%d\n", dp[ls%2][lt]); }    return 0;}

View Code

Poj1458--dp,lcs