hduoj1159  dp,lcs

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25570    Accepted Submission(s):
11363

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 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 Input

abcfbc abfcab
programming contest
abcd mnp

Sample Output

4
2
0

求lcs，水题，用了滚动数组才过

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=1000100;
char s[maxn],t[maxn];
int dp[2][maxn];

int main()
{
    while(scanf("%s%s",s,t)!=EOF){
        int ls=strlen(s),lt=strlen(t);
        memset(dp,0,sizeof(dp));
        char *ss=s-1,*tt=t-1;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=ls;i++){
            for(int j=1;j<=lt;j++){
                if(ss[i]==tt[j]) dp[i%2][j]=dp[(i+1)%2][j-1]+1;
                else dp[i%2][j]=max(dp[(i+1)%2][j],dp[i%2][j-1]);
            }
        }
        printf("%d\n",dp[ls%2][lt]);
    }
    return 0;
}