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Common Subsequence

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attack
发布2018-04-12 12:16:37
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发布2018-04-12 12:16:37
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文章被收录于专栏:数据结构与算法

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.

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

代码语言:javascript
复制
abcfbc         abfcab
programming    contest 
abcd           mnp

Sample Output

代码语言:javascript
复制
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

代码语言:javascript
复制
 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 }
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