hdu 1159 Common Subsequence

最新推荐文章于 2025-09-05 12:54:34 发布

原创最新推荐文章于 2025-09-05 12:54:34 发布 · 487 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#ACM #c++ #动态规划 #dp #子序列

动态规划专栏收录该内容

6 篇文章

订阅专栏

本文介绍了解决最长公共子序列问题的一种动态规划方法，并提供了一段完整的C++实现代码。通过对输入字符串进行处理，利用二维数组记录中间状态，最终找到两个字符串之间的最长公共子序列长度。

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18034 Accepted Submission(s): 7605

Problem 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, 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.

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

求最长公共子序列！

AC代码

#include <iostream>
using namespace std;
char a[1001],b[1001];
int dp[1001][1001];
int MAX(int a,int b)
{
	if(a>b)
		return a;
	else 
		return b;
}
int main()
{
	int i,j,n,m;
	while(scanf("%s %s",&a,&b)!=EOF)
	{
		n=strlen(a);
		m=strlen(b);
		memset(dp,0,sizeof(dp));
		for (i=1;i<=n;i++)
		{
			for(j=1;j<=m;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]);
			}
		}
		printf("%d\n",dp[n][m]);
		
	}
	return 0;
}

谢谢阅读！