Common Subsequence HDU - 1159（动态规划最长公共子序列）

最新推荐文章于 2021-07-30 19:45:57 发布

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本文介绍了一种寻找两个字符串中最长公共子序列的算法实现，通过动态规划的方法，有效地解决了该问题，并给出了具体的代码实现及样例输入输出。

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

思路：

理解下面图片的内容就可以AC啦！
在这里插入图片描述

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[10000],b[10000];
int dp[1000][1000];
int main()
{
    while(~scanf("%s %s",a+1,b+1))
    {
        memset(dp,0,sizeof(dp));
        int i,j;
        int l=strlen(a+1);
        int l1=strlen(b+1);
        for(i=1; i<=l; i++)
            dp[i][0]=0;
        for(j=1; j<=l1; j++)
            dp[0][j]=0;
        int maxx=0;
        for(i=1; i<=l; i++)
        {
            for(j=1; j<=l1; j++)
            {
                if(a[i]==b[j])
                    dp[i][j]=dp[i-1][j-1]+1;
                //加上下面的else语句表示求的是非连续的，删掉求出的则是连续的
                else 
                    dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                maxx=max(dp[i][j],maxx);//放在for循环内每次要更新
            }

        }
        printf("%d\n",maxx);
    }
    return 0;
}