HDU 1159 Common Subsequence

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最新推荐文章于 2024-08-22 20:10:34 发布

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本文详细介绍了如何求解两个字符串的最长公共子序列问题，包括问题定义、解决思路及C++实现代码。通过动态规划的方法，实现了高效查找最长公共子序列的功能。

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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

PS:最后一个dp[i][j]={dp[i-1][j-1];dp[i-1][j];dp[i][j-1]}三者之中比较是否相同;
最长公共子序列问题具有最优子结构性质
设
X = { x1 , ... , xm }
Y = { y1 , ... , yn }
及它们的最长子序列
Z = { z1 , ... , zk }
则
1、若 xm = yn ， 则 zk = xm = yn，且Z[k-1] 是 X[m-1] 和 Y[n-1] 的最长公共子序列
2、若 xm != yn ，且 zk != xm , 则 Z 是 X[m-1] 和 Y 的最长公共子序列
3、若 xm != yn , 且 zk != yn , 则 Z 是 Y[n-1] 和 X 的最长公共子序列
由性质导出子问题的递归结构
当 i = 0 , j = 0 时 , c[i][j] = 0
当 i , j > 0 ; xi = yi 时 , c[i][j] = c[i-1][j-1] + 1
当 i , j > 0 ; xi != yi 时 , c[i][j] = max { c[i][j-1] , c[i-1][j] }
#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int i,j,dp[500][500];
    string s1,s2;
    while(cin>>s1>>s2)
    {
        memset(dp,0,sizeof(dp));
        for(i=1; i<=s1.size(); i++)
        {
            for(j=1; j<=s2.size(); j++)
            {
                if(s1[i-1]==s2[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[s1.size()][s2.size()]<<endl;
    }
    return 0;
}