HDU - 1159-Common Subsequence

最新推荐文章于 2019-03-05 15:55:13 发布

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本文深入探讨了最长公共子序列(LCS)算法，详细解释了如何通过动态规划方法求解两个字符串的最长公共子序列长度。通过实例演示了算法的具体实现过程，包括输入输出示例，以及使用C++编程语言的代码实现。

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, xij = 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.
The program input is from a text file. Each data set in the file contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct. 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.
Input
abcfbc abfcab
programming contest
abcd mnp
Output
4
2
0
Sample Input
abcfbc abfcab
programming contest
abcd mnp
Sample Output
4
2
0

#include<bits/stdc++.h>
#include <iostream>
using namespace std;

int main()
{
	string a, b;
	while (cin >> a >> b)
	{
		
		int dp[1000][1000], o, p;
		memset(dp, 0, sizeof(dp));
		o = a.size();
		p = b.size();
		for (int i = 1; i <= o; i++)
		{
			for (int j = 1; j <= p; j++)
			{
				if (a[i - 1] == b[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1;
				else
				{
					dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
				}
			}
		}
		cout << dp[o][p] << endl;
	}

}