POJ 1458 / HDU 1159 / Southeastern Europe 2003 Common Subsequence (DP&LCS)

本文介绍了如何通过动态规划解决最大公共子序列问题，并提供了样例输入和输出以供参考。

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Common Subsequence

Time Limit: 1000MS

Memory Limit: 10000K

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

参考这篇文章。

完整代码：

/*POJ: 0ms,964KB*/
/*HDU: 31ms,2176KB*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1010;

char a[maxn], b[maxn];
int dp[maxn][maxn];

int main(void)
{
	while (~scanf("%s%s", a, b))
	{
		int lena = strlen(a), lenb = strlen(b);
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= lena; ++i)
			for (int j = 1; j <= lenb; ++j)
				dp[i][j] = (a[i - 1] == b[j - 1] ? dp[i - 1][j - 1] + 1 : max(dp[i - 1][j], dp[i][j - 1]));
		printf("%d\n", dp[lena][lenb]);
	}
	return 0;
}