动态规划 Common Subsequence

描述

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.

输入

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.

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

一道非常简单的动态规划题，菜鸡小白正在研究之中

#include <iostream>
#include <algorithm>
#include <string>
#define N 1001
#include <cstring>
using namespace std;
int a[N][N];
int main()
{
    int n, m, j, k, i;
    string   x, y;
    while (cin >> x >> y)
    {
        n = x.length();
        m = y.length();
        for (i = ; i < n; i++)
        {
            for (j = ; j < m; j++)
            {
                a[i][j] = ;
            }
        }
        for (i = ; i <= n; i++)
        {
            for (j = ; j <= m; j++)
            {
                if (x[i-] == y[j-])
                    a[i][j] = a[i - ][j - ] + ;
                else if (a[i - ][j] > a[i][j - ])
                    a[i][j] = a[i - ][j];
                else a[i][j] = a[i][j - ];
            }
        }
        cout << a[n][m] << endl;
    }
}