LCS pku1458

Common Subsequence

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 15634		Accepted: 6055

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, 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

LCS:
 问题描述

给定两个序列
X = { x1 , x2 , ... , xm }
Y = { y1 , y2 , ... , yn }
求X和Y的一个最长公共子序列

举例
X = { a , b , c , b , d , a , b }
Y = { b , d , c , a , b , a }
最长公共子序列为
LSC = { b , c , b , a }

二 问题分析

分析：

最长公共子序列问题具有最优子结构性质

设
X = { x1 , ... , xm }
Y = { y1 , ... , yn }
及它们的最长子序列
Z = { z1 , ... , zk }

则

1、若 xm = yn ， 则 zk = xm = yn，且Z[k-1] 是 X[m-1] 和 Y[n-1] 的最长公共子序列
2、若 xm != yn ，且 zk != xm , 则 Z 是 X[m-1] 和 Y 的最长公共子序列
3、若 xm != yn , 且 zk != yn , 则 Z 是 Y[n-1] 和 X 的最长公共子序列

由性质导出子问题的递归结构

当 i = 0 , j = 0 时 , c[i][j] = 0
当 i , j > 0 ; xi = yi 时 , c[i][j] = c[i-1][j-1] + 1
当 i , j > 0 ; xi != yi 时 , c[i][j] = max { c[i][j-1] , c[i-1][j] }

该题的代码：

#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)
{
    memset(dp, 0, sizeof(dp));
    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)));
    }
}