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UVA10405

发布时间:2022-07-27 13:56:10

UVA - 10405

Longest Common Subsequence
Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu

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Description

 

Problem C: Longest Common Subsequence

Sequence 1:                

Sequence 2:                
Given two sequences of characters, print the length of the longest common subsequence of both sequences. For example, the longest common subsequence of the following two sequences:

abcdgh
aedfhr
is adh of length 3.

Input consists of pairs of lines. The first line of a pair contains the first string and the second line contains the second string. Each string is on a separate line and consists of at most 1,000 characters

For each subsequent pair of input lines, output a line containing one integer number which satisfies the criteria stated above.

Sample input

a1b2c3d4e
zz1yy2xx3ww4vv
abcdgh
aedfhr
abcdefghijklmnopqrstuvwxyz
a0b0c0d0e0f0g0h0i0j0k0l0m0n0o0p0q0r0s0t0u0v0w0x0y0z0
abcdefghijklmnzyxwvutsrqpo
opqrstuvwxyzabcdefghijklmn

Output for the sample input

4
3
26
14




#include <cstdlib>
#include <cstring>
#include <string>
#include <iostream>
using namespace std;
#define N 10005
string str1,str2;
int dp[N+1][N+1] ;
int maxx(int a , int b)
{     
    if(a > b) return a ;  
    return b ;
}
int LCSL(int len1 , int len2)
{     
    int i , j ;   
    int len = maxx(len1 , len2);    
    for( i = 0 ; i <= len; i++ )    
        dp[i][0] = 0,dp[0][i] = 0;   
    for( i = 1 ; i<= len1 ; i++)      
        for( j = 1 ; j <= len2 ; j++)      
            if(str1[i - 1] == str2[j - 1])           
                dp[i][j] = dp[i - 1][ j - 1] + 1 ;      
            else                 
                dp[i][j] = maxx(dp[i - 1][ j ] , dp[i][j - 1]) ;    
            return dp[len1][len2]; }
int main()
{     
    while(getline(cin,str1))
    {          
        getline(cin,str2);      
        int len1=str1.size();      
        int len2 =str2.size();   
        cout<<LCSL(len1 , len2)<<endl;  
    }     
    return 0;
}



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