poj1458-Common Subsequence（最长公共子序列，LCS）

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原创最新推荐文章于 2026-02-24 01:08:14 发布 723 阅读 AI 写同款· GEO 优化 ›

CMOS Sensor的调试经验分享我这里要介绍的就是CMOS摄像头的一些调试经验。首先，要认识CMOS摄像头的结构。我们通常拿到的是集成封装好的模组，一般由三个部分组成：镜头、感应器和图像信号处理器构成。一般情况下，集成好的模组我们只看到外面的镜头、接口和封装壳，这种一般是固定焦距的。有些厂商只提供芯片，需要自己安装镜头，镜头要选择合适大小的镜头，如果没有夜视要求的话，最好选择带有红外滤光的镜头，因为一般的sensor都... 阅读详情

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 43586		Accepted: 17731

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

题意：给出两个字符串，找出它们的最长公共子序列

这个题主要是考察LCS（最长公共子序列）部分内容，关键是明白LCS的原理，不明白的来参考我另一篇博客

南阳OJ36-最长公共子序列（LCS）

只要找出模板就行了！

下面给出代码！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char str1[1010],str2[1010];
int dp[1010][1010];
int main()
{
	while(scanf("%s%s",str1,str2)!=EOF)
	{
		memset(dp,0,sizeof(dp));
		int len1=strlen(str1);
		int len2=strlen(str2);
		for(int i=1;i<=len1;i++)
		{
			for(int j=1;j<=len2;j++)
			{
				if(str1[i-1]==str2[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]);
	}
	return 0;
}