Common Subsequence

最新推荐文章于 2020-09-22 23:59:10 发布

yuebaba

最新推荐文章于 2020-09-22 23:59:10 发布

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分类专栏： lcs+lis

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本文介绍了一种解决最长公共子序列问题的算法实现。通过动态规划方法，该程序能够找出两个字符串序列间的最长公共子序列长度。输入为两组字符串，输出其最长公共子序列的长度。

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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 ij = 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

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn=1e3;
int dp[maxn][maxn];
int main()
{
	char a[maxn],b[maxn];
	while(scanf("%s %s",a,b)!=EOF)
	{
		int a_l=strlen(a),b_l=strlen(b);
		memset(dp,0,sizeof(dp));
		for(int i=0;i<a_l;i++)
		for(int j=0;j<b_l;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[a_l][b_l]<<endl;
	}
	return 0;
}