公共子序列(common subsequence) boj1134

Common SubsequenceSubmit: 407   Accepted:166Time Limit: 1000MS  Memory Limit: 65536KDescription
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.


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



最长公共子序列

O(n^2)时间复杂度，DP

公式如下,其中A,B表示两个字符串，c[i][j]存储的是中间结果

c[i][j]=0,                         i=0,j=0
c[i][j]=c[i-1][j-1]+1,             A[i]==B[j]
c[i][j]=max(c[i-1][j],c[i][]j-1)   A[i]!=B[j]

c[strlen(A)][strlen(B)]即为LCS的结果



http://blog.csdn.net/fangxia722/archive/2008/10/10/3051524.aspx

代码：

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

#define MAX(a, b) ((a) > (b) ? (a) : (b))

char A[1000], B[1000];
int dp[1000][1000];

int LCS(char *a, char *b, int a_len, int b_len)
{
    int i, j;
    for (i=1 ; i<=a_len ; i++)
    {
        for (j=1 ; j<=b_len ; j++)
        {
            if (a[i-1] == b[j-1])
            {
                dp[i][j] = dp[i-1][j-1] + 1;
            }
            else
            {
                dp[i][j] = MAX(dp[i-1][j], dp[i][j-1]);
            }
        }
    }
    return dp[a_len][b_len];
}

int main(int argc, char *argv[])
{
    while (EOF != scanf("%s%s", A, B))
    {
        printf("%d\n", LCS(A, B, strlen(A), strlen(B)));
    }
}