HDU 1159 Common Subsequence

Common Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30912    Accepted Submission(s): 13987

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

 

 

Recommend

Ignatius

 

 

L[i+1][j+1]=max(L[i][j+1],L[i+1][j]); L[i][j]表示字符串A的1~i+1子串与字符串B的1~j+1子串的公共子串的长度。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn=1050;
int L[maxn][maxn];
char s1[maxn];
char s2[maxn];

int main()
{
    int len1,len2;
    while(scanf("%s%s",s1,s2)==2)
    {
        len1=strlen(s1);
        len2=strlen(s2);
        for(int i=0;i<len1;i++)
            for(int j=0;j<len2;j++)
                L[i][j]=0;
        for(int i=0;i<len1;i++)
        {
            for(int j=0;j<len2;j++)
                if(s1[i]==s2[j])  L[i+1][j+1]=L[i][j]+1;
                else  L[i+1][j+1]=max(L[i][j+1],L[i+1][j]);
        }
        printf("%d\n",L[len1][len2]);
    }
}