HDU - 1159 Common Subsequence （最长公共子序列，DP）

原创于 2019-01-30 09:30:24 发布 · 192 阅读

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本文探讨了最长公共子序列(LCS)问题，这是一种经典的计算机科学问题，常见于字符串比较和生物信息学等领域。文章详细介绍了如何使用动态规划解决LCS问题，并提供了一个C++实现示例。

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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
问题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1159
问题简述：输入两串字符串，求其中最长公共子序列
问题分析：很明显的DP题啦。状态转移方程：if(p1[i-1]==p2[i-1]) dp[i][j]=dp[i-1][j-1]+1
else dp[i][j]=max(dp[i-1][j],dp[i][j-1])
我觉得这篇博客讲的挺好：https://blog.youkuaiyun.com/u010579068/article/details/49207347
AC通过的C++语言程序如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include <algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
const int m = 1000;
int dp[m + 5][m + 5];
char p1[m];
char p2[m];
void DP(int x,int y)
{
	for (int i = 1; i <= x; i++)
	{
		for (int j = 1; j <= y; j++)
		{
			if (p1[i - 1] == p2[j - 1]) { dp[i][j] = dp[i - 1][j - 1] + 1; }
			else { dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); }
		}
	}
	return;
}
int main()
{
	while (cin >> p1 >> p2)
	{
		memset(dp, NULL, sizeof(dp));
		int len1 = strlen(p1);
		int len2 = strlen(p2);
		DP(len1, len2);
		cout << dp[len1][len2] << endl;
	}
	return 0;
}