joj1968: Common Subsequence

 1968: Common Subsequence

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	1s	4096K	781	227	Standard

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, x_ij = 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.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

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

 

 

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 经典的DP，最大公共子序列。

 

#include <stdio.h>
#include <string.h>

int dp[500][500];
char c1[500], c2[500];

int main()
{
    //freopen("in.txt", "r", stdin);

    while (scanf("%s%s", c1, c2) == 2)
    {
        int len1, len2, i, j;
        len1 = strlen(c1);
        len2 = strlen(c2);

        for (i = 0; i <= len1; ++i)
        {
            dp[0][i] = 0;
        }
        for (j = 0; j <= len2; ++j)
        {
            dp[j][0] = 0;
        }

        for(i = 1; i <= len1; i++)
        {
            for(j = 1; j <= len2; j++)
            {
                if(c1[i-1] == c2[j-1])
                {
                    dp[i][j] = dp[i-1][j-1] + 1;
                }
                else
                {
                    if(dp[i-1][j] > dp[i][j-1])
                    {
                        dp[i][j] = dp[i-1][j];
                    }
                    else
                    {
                        dp[i][j] = dp[i][j-1];
                    }
                }
            }
        }
        
        printf("%d\n", dp[len1][len2]);
    }
    
    return 0;
}