Problem B

Problem 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.
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. 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

题意：给你两个字符数列，求相似度（最长公共子序列）

解题思路：动态规划，看了一下午课件没找到最长公共子序列的例题，在一个博客上看到了动态规划的讲解，刚开始还不大明白状态转移方程是怎么回事，，照着写了一遍，看看图对着代码走了一遍就明白了，在二维数组中查找相同的元素，DP[i][j]记录当前状态的最长公共子序列的长度，另外如果用一个flag[i][j]记录状态转移的过程，就可以输出最长公共子序列；

感悟：动态规划太抽象了，但是想出来了，就很好谢了；

代码：

#include

#include

#include

#define maxn 1005

using namespace std;

char s1[maxn],s2[maxn];

int dp[maxn][maxn];//记录当前状态的最长子序列长度

int main()

{

    //freopen("in.txt","r",stdin);

    while(~scanf("%s",&s1))

    {

        scanf("%s",&s2);

        for(int i=1;i<=strlen(s1);i++)

        {

            for(int j=1;j<=strlen(s2);j++)

            {

                if(s1[i-1]==s2[j-1])

                    dp[i][j]=dp[i-1][j-1]+1;

                else if(dp[i][j-1]>dp[i-1][j])

                    dp[i][j]=dp[i][j-1];

                else

                    dp[i][j]=dp[i-1][j];

            }

        }

        printf("%d\n",dp[strlen(s1)][strlen(s2)]);

    }

}