HDU 1159 Common Subsequence(최대 공통 하위 시퀀스 동적 계획)

Common Subsequence

Problem Description
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = another sequence Z = is a subsequence of X if there exists a strictly increasing sequence of indices of X such that for all j = 1,2,...,k, xij = zj. For example, Z = is a subsequence of X = 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.

Sample Input
abcfbc abfcab programming contest abcd mnp

Sample Output
4 2 0

코드:

 1 #include<iostream>

 2 #include<cstring>

 3 #define max(a,b) (a>b?a:b)

 4 #define MAXN 1001+10

 5 using namespace std;

 6 

 7 int c[MAXN][MAXN];

 8 

 9 int main()

10 {

11     //freopen("in.txt","r",stdin);

12     int s1_len,s2_len,i,j,t;

13     char s1[MAXN],s2[MAXN];

14     cin>>t;

15     while(t--)

16     {

17         cin>>s1>>s2;

18         s1_len=strlen(s1);

19         s2_len=strlen(s2);

20         for(i=0;i<=s1_len;i++)

21         c[i][0]=0;

22         for(i=0;i<=s2_len;i++)

23         c[0][i]=0;

24         for(i=1;i<=s1_len;i++)

25         for(j=1;j<=s2_len;j++)

26         if(s1[i-1]==s2[j-1])

27         c[i][j]=c[i-1][j-1]+1;

28         else

29         c[i][j]=max(c[i-1][j],c[i][j-1]);

30         cout<<c[s1_len][s2_len]<<endl;

31     }

32     return 0;

33 }

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POJ 2442 Sequence

Sequence Time Limit: 6000MS Memory Limit: 65536K Total Submissions: 6120 Accepted: 1897 Description Given m sequences, e...

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HDU 1159 Common Subsequence(최대 공통 하위 시퀀스 동적 계획)

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