HDU1159 : Common Subsequence（LCS）

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41128    Accepted Submission(s): 18995

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
  
    
    
    
    
  
    
    
    
    


  
    
    
    
    
  
    
    
    
    

   
     
     
     
     
#include
#include
#include
using namespace std;

const int maxn = 1000 ;

char s1[maxn] , s2[maxn] ;
int dp[maxn][maxn] ;
int len1 , len2 ;

int main() {
        ios :: sync_with_stdio(false) ;
	while(cin >> s1 >> s2) {
		len1 = strlen(s1) ; len2 = strlen(s2) ;
		memset(dp , 0 , sizeof(dp)) ;
		for(int i = 1 ; i <= len1 ; i ++) {
			for(int j = 1 ; j <= len2 ; j ++) {
				if(s1[i - 1] == s2[j - 1]) dp[i][j] = dp[i - 1][j - 1] + 1 ;
				else dp[i][j] = max(dp[i - 1][j] , dp[i][j - 1]) ;
			}
		}
		cout << dp[len1][len2] << endl ;
	}
	return 0 ;
}