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

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#poj1458 #Common Subsequence #最长公共子序列 #LCS

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本文介绍了一种解决最长公共子序列(LCS)问题的方法，通过动态规划算法高效地找出两个字符串之间的最长公共子序列长度，并附带了完整的C++实现代码。

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;
}