Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25383 Accepted Submission(s): 11253

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

Source Southeastern Europe 2003

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1159

題目大意：求最長公共子序列(LCS)

題目分析：LCS模板題

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int const MAX = 500;
char s1[MAX], s2[MAX];
int dp[MAX][MAX];

int main()
{
    while(scanf("%s %s", s1, s2) != EOF)
    {
        memset(dp, 0, sizeof(dp));
        int l1 = strlen(s1);
        int l2 = strlen(s2);
        for(int i = 1; i <= l1; i++)
        {
            for(int j = 1; j <= l2; j++)
            {
                if(s1[i - 1] == s2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                else
                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        printf("%d\n", dp[l1][l2]);
    }
}

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