Common Subsequence HDU - 1159 （最大公共子序列）-优快云博客

本文介绍了一种使用滚动数组优化空间复杂度的最长公共子序列（LCS）算法实现，通过对比两个字符串序列，找出它们之间的最长公共子序列长度。

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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.

Input

abcfbc abfcab
programming contest 
abcd mnp

Output

4
2
0

Sample Input

abcfbc abfcab
programming contest 
abcd mnp

Sample Output

4
2
0

题目大意与思路分析：

就是给两个字符串求他们的公共子序列。

我运用了滚动数组节省空间，dp数组记忆化搜索，如果a[i]==b[j]，此时就要再上一层对应位置+1；如果不相等则是看本层前一个所对应的dp数组与上一层对应位置的dp数组值判断大小取最大。最后记得比较dp[0][l2],dp[1][l2]因为是滚动数组所以不知道哪一个是最大的，所以做出最后一步的判断，输出即可。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[1200],b[1200];    //定义两个字符串数组
int dp[3][12200];        //定义dp数组
int main()
{
    while(~scanf("%s %s",a,b))
    {
        memset(dp,0,sizeof(dp));
        int l1=strlen(a);
        int l2=strlen(b);
        for(int i=0; i<l1; i++)
            for(int j=0; j<l2; j++)
            {
                if(a[i]==b[j])  //如果相等，则是在前一层的基础上+1
                    dp[(i+1)%2][j+1]=dp[i%2][j]+1;
                else            //若不相等，则比较这一层的前一个与上一层同位置判断求出最大
                    dp[(i+1)%2][j+1]=max(dp[i%2][j+1],dp[(i+1)%2][j]);
            }
        printf("%d\n",max(dp[1][l2],dp[0][l2]));//用了滚动数组，所以要比较dp下标0与1的大小
    }
    return 0;
}