HDU 1159 dp+最长公共子序列

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本文介绍了解决HDU1159问题——寻找两个字符串的最长公共子序列的经典动态规划算法。通过使用二维数组dp记录状态，实现高效求解。输入为两字符串，输出为最大公共子序列长度。

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HDU 1159
Common Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51579 Accepted Submission(s): 23755

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

问题分析：
此问题是经典的dp问题，公共序列可以用一个二维数组dp[i][j]保存每个点时的最大数字。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
using namespace std; 
 char a[1000],b[1000];int dp[1000][1000];
int main()
{
  
   while(cin>>a>>b)
   {
   	memset(dp,0,sizeof(dp));
   int len1,len2;
   len1=strlen(a);len2=strlen(b);
   for(int i=1;i<=len1;i++)
   for(int j=1;j<=len2;j++)
   {
   	if(b[j-1]==a[i-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[len1][len2]<<endl;
}
}