HDUOJ--1159Common Subsequence

 

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19595    Accepted Submission(s): 8326

Problem 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.
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

 

 

Source

Southeastern Europe 2003

简单的lcs

lcs 模板：

int lcs(const char *a,const char *b)

{

    int i,j;

    int m=strlen(a),n=strlen(b);

    mar[0][0]=0;

    for(i=1;i<=m;i++)

        mar[i][0]=0;

    for(i=1;i<=n;i++)

        mar[0][i]=0;

    for(i=1;i<=m;i++)

    {

        for(j=1;j<=n;j++)

        {

            if(a[i-1]==b[j-1])

                mar[i][j]=mar[i-1][j-1]+1;

            else

                mar[i][j]=mar[i-1][j]>mar[i][j-1]?mar[i-1][j]:mar[i][j-1];

        }

    }

    return mar[m][n];

}

View Code

此题的代码：

 1 #include<stdio.h>

 2 #include<string.h>

 3 #define maxn 1000

 4 int mar[maxn][maxn];

 5 char x[maxn],y[maxn];

 6 

 7 int lcs(const char *a,const char *b)

 8 {

 9     int i,j;

10     int m=strlen(a),n=strlen(b);

11     mar[0][0]=0;

12     for(i=1;i<=m;i++)

13         mar[i][0]=0;

14     for(i=1;i<=n;i++)

15         mar[0][i]=0;

16     for(i=1;i<=m;i++)

17     {

18         for(j=1;j<=n;j++)

19         {

20             if(a[i-1]==b[j-1])

21                 mar[i][j]=mar[i-1][j-1]+1;

22             else

23                 mar[i][j]=mar[i-1][j]>mar[i][j-1]?mar[i-1][j]:mar[i][j-1];

24         }

25     }

26     return mar[m][n];

27 }

28 int main()

29 {

30     while(scanf("%s%s",&x,&y)!=EOF)    

31      printf("%d\n",lcs(x,y));

32     return 0;

33 }

View Code