POJ 1458 Common Subsequence（最长公共子序列问题）

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本文介绍了一种解决最长公共子序列问题的经典算法，并提供了两种实现方式：一种使用二维动态规划表，另一种通过优化空间复杂度减少内存消耗。适用于对算法和数据结构有一定了解的读者。

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Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions:56465		Accepted: 23531

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

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
int dp[1002][1002];
using namespace std;
int main()
{
	char a[1002], b[1002];
	while (scanf("%s%s",&a,&b)!=EOF)
	{
		int L1 = strlen(a), L2 = strlen(b);
		memset(dp, 0, sizeof(dp));
		for (int i = L1; i >= 1; i--)
			a[i] = a[i - 1];
		for (int i = L2; i >= 1; i--)
			b[i] = b[i - 1];
		for (int i = 1; i <= L1; i++)
		{
			for (int j = 1; j <= L2; j++)
			{
					dp[i][j] = max(dp[i][j-1],dp[i-1][j]);
					if (a[i] == b[j])
						dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
			}
		}
		printf("%d\n", dp[L1][L2]);
	}
	return 0;
}

优化空间复杂度

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
int dp[2][1002];
using namespace std;
int main()
{
	char a[1002], b[1002];
	while (scanf("%s%s",&a,&b)!=EOF)
	{
		int L1 = strlen(a), L2 = strlen(b);
		memset(dp, 0, sizeof(dp));
		for (int i = L1; i >= 1; i--)
			a[i] = a[i - 1];
		for (int i = L2; i >= 1; i--)
			b[i] = b[i - 1];
		for (int i = 1; i <= L1; i++)
		{
			for (int j = 1; j <= L2; j++)
			{
					dp[i&1][j] = max(dp[i&1][j-1],dp[(i-1)&1][j]);
					if (a[i] == b[j])
						dp[i&1][j] = max(dp[i&1][j], dp[(i - 1)&1][j - 1] + 1);
			}
		}
		printf("%d\n", max(dp[0][L2],dp[1][L2]));
	}
	return 0;
}