HDU 1159 Common Subsequence（dp基础：最长公共子序列）

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本文介绍了一种解决最长公共子序列问题的算法，并通过一个C++程序实例展示了如何求解两个字符串之间的最大长度公共子序列。该算法利用动态规划的方法，通过构建二维数组来记录中间结果。

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Common Subsequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46318 Accepted Submission(s): 21236

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

//输出是abca   
#include<stdio.h>  
#include<string.h>  
#include<string>  
#include<iostream>  
#include<algorithm>  
using namespace std;  
int dp[1005][1005];  
int main(){  
    string a,b,c;
	while(cin>>a>>b)
	{
		for(int i=0;i<=a.length();i++)
		{  
	        for(int j=0;j<=b.length();j++)
			{  
	            if(a[i]==b[j])  
	                dp[i+1][j+1]=dp[i][j]+1;  
	            else dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1]);  
	        }  
    	}    
    	cout<<dp[a.length()][b.length()]<<endl; 
	}  
    return 0;  
}