Common Subsequence

最新推荐文章于 2021-03-13 16:53:17 发布

weixin_30432007

最新推荐文章于 2021-03-13 16:53:17 发布

阅读量46

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zwfymqz/p/7202106.html

本文介绍了一种使用动态规划解决最长公共子序列问题的方法。通过分析两个字符串序列，找到它们之间的最长公共子序列长度。代码示例展示了如何实现这一算法，并提供了一个简单的输入输出示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 53247		Accepted: 22084

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

Source

Southeastern Europe 2003

用dp[n][m]表示第一个喘选到n，第二个串选到m

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;
 7 const int MAXN=601;
 8 const int maxn=0x7fffff;
 9 char a[MAXN];
10 char b[MAXN];
11 int dp[MAXN][MAXN];
12 int la,lb;
13 int main()
14 {
15     while(~scanf("%s%s",a,b))
16     {
17         la=strlen(a);
18         lb=strlen(b);
19         memset(dp,0,sizeof(dp));
20         //dp[0][0]=1;
21         for(int i=1;i<=la;i++)
22         {
23             for(int j=1;j<=lb;j++)
24             {
25                 
26                 if(a[i-1]==b[j-1])
27                     dp[i][j]=max(dp[i][j],dp[i-1][j-1]+1);
28                 else dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
29             }
30         }
31         printf("%d\n",dp[la][lb]);
32     }
33     return 0;
34 }