总时间限制: 1000ms 内存限制: 65536kB

描述

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x, x, …, x > another sequence Z = < z, z, …, z > is a subsequence of X if there exists a strictly increasing sequence < i, i, …, i > of indices of X such that for all j = 1,2,…,k, x = 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.

输入

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.

输出

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.

样例输入

abcfbc         abfcab
programming    contest 
abcd           mnp

样例输出

4
2
0

思路

输入串s1, s2, 设 MaxLen(i, j) 表示:

s1左边第i个字符形成的字串, 与s2左边第j个字符形成的字串的最长公共子序列的长度(i, j从0开始计数).

MaxLen(i, j)就是本题的状态.

假定 len1 = strlen(s1), len2 = strlen(s2) , 那么题目要求MaxLen(s1, s2).

显然, 边界条件为:

• MaxLen(n, 0) = 0(n = 0, …, len1)
• MaxLen(0, n) = 0(n=0, …, len2)

递推式为:

if( s1[i-1] == s2[j-1] )
    MaxLen(i, j) = MaxLen(i-1, j-1) + 1;
else
    MaxLen(i, j) = Max( MaxLen(i, j-1), MaxLen(i-1, j) );

代码

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
char s1[1024];
char s2[1024];
int maxLen[1024][1024];
int main() {
    while(cin >> s1 >> s2) {
        int len1 = strlen(s1);
        int len2 = strlen(s2);
        for(int i = 0; i <= len1; i++) 
            maxLen[i][0] = 0;
        for(int i = 0; i <= len2; i++) 
            maxLen[0][i] = 0;
        /* Compute every state */
        for(int i = 1; i <= len1; i++) {
            for(int j = 1; j <= len2; j++) {
                if( s1[i-1] == s2[j-1] )
                    maxLen[i][j] = maxLen[i-1][j-1] + 1;
                else
                    maxLen[i][j] = max(maxLen[i][j-1], 
                            maxLen[i-1][j]);
            }
        }
        cout << maxLen[len1][len2] << endl;
    }
    return 0;
}