POJ 1458 Common Subsequence (最长公共子序列)

Common Subsequence

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, 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

 

典型的动态规划问题，求最长公共子序列(LCS). 这篇文章最长公共子序列(LCS)已经作出了相关解释。

下面直接给出代码：

#include <iostream>
#include <string.h>
using namespace std;
//LCS问题
int main()
{
    string s;
    string t;
    while(cin>>s>>t)
    {
        const char* a = s.c_str();
        const char* b = t.c_str();
        int m = strlen(a) + 1;
        int n = strlen(b) + 1;
        int** c = new int*[m];
        for(int i = 0; i < m; i++)
            c[i] = new int[n];

        //m行n列
        for(int i = 0; i < m; i++)
            c[i][0] = 0;
        for(int i = 0; i < n; i++)
            c[0][i] = 0;

        for(int i = 1; i < m; i++)
            for(int j = 1; j < n; j++)
            {
                if(a[i-1] == b[j-1])
                    c[i][j] = c[i-1][j-1] + 1;
                else if(c[i][j-1] >= c[i-1][j])
                    c[i][j] = c[i][j-1];
                else
                    c[i][j] = c[i-1][j];
            }

        cout<<c[m-1][n-1]<<endl;
        for(int i = 0; i < m; i++)
            delete c[i];
        delete c;
        
    }
    return 0;
}