POJ 题目1458 Common Subsequence（最长公共子序列长度纯模板）

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本文深入探讨了深度学习在自然语言处理领域的应用，包括词法分析、语义理解、文本生成等关键技术，以及如何利用这些技术进行高效的数据处理与分析。

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Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 39748		Accepted: 15983

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, 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

Source

Southeastern Europe 2003

好无聊啊，水道题爽两把

ac代码

#include<stdio.h>
#include<string.h>
#define max(a,b) (a>b?a:b)
int dp[1010][1010];
int main()
{
	char s1[10010],s2[100010];
	while(scanf("%s%s",s1,s2)!=EOF)
	{
		int i,j,k,len1,len2;
		memset(dp,0,sizeof(dp));
		len1=strlen(s1);
		len2=strlen(s2);
		for(i=1;i<=len1;i++)
		{
			for(j=1;j<=len2;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[len1][len2]);
	}
}