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最新推荐文章于 2021-03-07 15:01:42 发布

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本文介绍了一种求解两个字符串最长公共子序列问题的动态规划算法实现，并提供了完整的C++代码示例。该算法通过构建二维DP数组来记录两个序列间的最长公共子序列长度。

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

<span style="font-family:Courier New;">#include <iostream>
#include <string>
using namespace std;

const int MAXN=1000;
int dp[MAXN][MAXN];

int main()
{
    string s1,s2;
    while(cin>>s1>>s2)
    {
        int Len1=s1.length();
        int Len2=s2.length();
        for(int j=0;j<=Len1;j++)
            dp[0][j]=0;
        for(int i=0;i<=Len2;i++)
            dp[i][0]=0;
        for(int i=1;i<=Len2;i++)
        {
            for(int j=1;j<=Len1;j++)
            {
                if(s1[j-1]==s2[i-1])
                    dp[i][j]=dp[i-1][j-1]+1;
                else
                    dp[i][j]=max(dp[i][j-1],dp[i-1][j]);
            }
        }
        cout<<dp[Len2][Len1]<<endl;
        s1.clear();
        s2.clear();
    }
    return 0;
}</span>