Common Subsequence - POJ 1458 最长非连续公共子序列

最新推荐文章于 2022-07-02 02:16:12 发布

原创最新推荐文章于 2022-07-02 02:16:12 发布 · 583 阅读

0 ·

CC 4.0 BY-SA版权

dp动态规划和递推同时被 2 个专栏收录

281 篇文章

订阅专栏

POJ

206 篇文章

订阅专栏

本文深入探讨了最长非连续公共子序列问题的解决方法，通过动态规划技术求解，提供了一种高效的算法解决方案，并附有AC代码实例。

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 37262		Accepted: 14880

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, x_{i_j} = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

题意：最长非连续公共子序列的长度。

思路：dp[i][j]表示到s1的i位置和s2的j位置的时候，最长非连续公共子序列的长度。dp[i][j]=max(dp[i-1][j],dp[i][j-1]);如果ij位置的字母相同的时候dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s1[1010],s2[1010];
int dp[1010][1010];
int main()
{ int i,j,k,len1,len2;
  while(~scanf("%s %s",s1,s2))
  { len1=strlen(s1);
    len2=strlen(s2);
    memset(dp,0,sizeof(dp));
    for(i=1;i<=len1;i++)
     for(j=1;j<=len2;j++)
     { dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
       if(s1[i-1]==s2[j-1])
        dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
     }
    printf("%d\n",dp[len1][len2]);
  }
}