3002

3002

Problem Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = &lt;x1, x2, ..., xm&gt; another sequence Z = &lt;z1, z2, ..., zk&gt; is a subsequence of X if there exists a strictly increasing sequence &lt;i1, i2, ..., ik&gt; of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = &lt;a, b, f, c&gt; is a subsequence of X = &lt;a, b, c, f, b, c&gt; with index sequence &lt;1, 2, 4, 6&gt;. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y. <br>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. <br>

 

Sample Input

   
   
   
   

    
    
    
    abcfbc abfcab
programming contest 
abcd mnp

   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    4
2
0

   
   
   
   

 

题意：

输入两个字符串，求他们的最长公共子序列的长度

思路：

第一反应是用搜索做，但后来想了一下，既然是在学动态规划，那肯定是用动态规划更好了。

首先用两个数组存放数据元素，然后对这两个数组进行比较。

如果序列对应的字符相同，如果序列对用的字符不同，则数组往后滚动。

AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
 
char s1[1000],s2[1000];
int dp[1000][1000];
int len1,len2;
 
void LCS()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    for(i = 1; i<=len1; i++)
    {
        for(j = 1; j<=len2; 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]);
        }
    }
}
 
int main()
{
    while(~scanf("%s%s",s1,s2))

{
        len1 = strlen(s1);
        len2 = strlen(s2);
        LCS();
        printf("%d\n",dp[len1][len2]);
    }
 
    return 0;
}