tzoj1510 Common Subsequence (LCS template longest common subsequence)

Time limit (Normal / Java): 1000MS / 10000MS memory limit: 65536KByte

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. 

Entry

 

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.

 

Export

 

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 title

Southeastern Europe 2003

 

 

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int N = 1000;
 6 char a[N],b[N];
 7 int dp[N][N];
 8 int main()
 9 {
10     int lena,lenb,i,j;
11     while(scanf("%s%s",a,b)!=EOF)
12     {
13         memset(dp,0,sizeof(dp));
14         lena=strlen(a);
15         lenb=strlen(b);
16         for(i=1;i<=lena;i++)
17         {
18             for(j=1;j<=lenb;j++)
19             {
20                 if(a[i-1]==b[j-1])
21                 {
22                     dp[i][j]=dp[i-1][j-1]+1;
23                 }
24                 else
25                 {
26                     dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
27                 }
28             }
29         }
30         printf("%d\n",dp[lena][lenb]);
31     }
32     return 0;
33 }