（最长公共子序列暴力） Common Subsequence (poj 1458)

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原文链接：http://www.cnblogs.com/YY56/p/5444205.html

本文介绍了一个经典的计算机科学问题——寻找两个字符串序列的最长公共子序列，并提供了一段C++代码实现。该程序从标准输入读取两组字符串，通过动态规划算法计算并输出它们的最长公共子序列长度。

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http://poj.org/problem?id=1458

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_{i_j} = 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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

#define met(a,b) (memset(a,b,sizeof(a)))
#define N 1100
#define INF 0xffffff

char s1[N], s2[N];
int dp[N][N];

int main()
{
    while(scanf("%s%s", s1, s2)!=EOF)
    {
        int i, j, len1=strlen(s1), len2=strlen(s2);

        met(dp, 0);
        
        ///要考虑下下标越界的问题
        for(i=1; i<=len1; i++)
        for(j=1; j<=len2; j++)
        {
            if(s1[i-1]==s2[j-1])
                dp[i][j] = dp[i-1][j-1] + 1;
            else dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
        }

        printf("%d\n", dp[len1][len2]);
    }
    return 0;
}