HDU OJ Common Subsequence

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原文链接：http://www.cnblogs.com/lzmfywz/articles/2355968.html

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#java

本文介绍了一种解决两个字符串最长公共子序列问题的方法，通过动态规划算法找到最优解。

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Common Subsequence

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 14 Accepted Submission(s) : 7

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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

分析：求两个字符串的最长公共子序列。

代码如下：

#include <iostream>
#include <cstring>
using namespace std;

char str1[1005];
char str2[1005];
int dp[1005][1005];

int main()
{
    int i,j;
    int Max;
    int len1,len2;
    while(cin>>str1>>str2)
    {
        len1=strlen(str1);
        len2=strlen(str2);
        for(i=0;i<len1;i++)
        {
            dp[i][0]=0;
        }
        for(i=0;i<len2;i++)
        {
            dp[0][i]=0;
        }
        for(i=1;i<=len1;i++)
        {
            for(j=1;j<=len2;j++)
            {
                if(str1[i-1]==str2[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[len1][len2]<<endl;
    }
    return 0;
}