Common Subsequence C++

最新推荐文章于 2024-07-27 13:09:04 发布

Zero_979

最新推荐文章于 2024-07-27 13:09:04 发布

阅读量304

点赞数

CC 4.0 BY-SA版权

分类专栏： dp 文章标签： ACM

本文链接：https://blog.youkuaiyun.com/Zero_979/article/details/80600852

dp 专栏收录该内容

39 篇文章

订阅专栏

本文介绍了解决最长公共子序列问题的一种动态规划方法。通过两组字符串输入，利用动态规划算法找出它们之间的最长公共子序列长度。示例代码展示了如何实现这一过程，并通过样例输入说明了其工作原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

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, x _{i_j} = 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         abfcab
programming    contest 
abcd           mnp

Sample Output

4
2
0

思路：

最长公共子序列问题。

代码：

#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int main()
{
    int n,m;
    string a,b;
    while(cin>>a>>b)
    {
        int dp[500][500];
        memset(dp,0,sizeof(dp));
        n=a.length();
        m=b.length();
        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;
                else dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j]);
            }
        }
        cout<<dp[n][m]<<endl;
    }
    return 0;
}