最长公共子序列(LCS) Easy HDU1599

题目：

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.

输入：

abcfbc abfcab
programming contest 
abcd mnp

输出：

4
2
0

找到两个字符串中最长的公共子序列，动态规划，具体过程分析见：最长公共子序列

主要是找到状态转移方程，

下面贴出代码：

#include <bits/stdc++.h>

using namespace std;

int main()
{
    char a[500],b[500];
    int c[500][500];
    int alen,blen;
    char *x=a+1,*y=b+1;
    while(cin>>x>>y)
    {
        memset(c,0,sizeof(c));
        alen=strlen(x);
        blen=strlen(y);
        for(int i=1;i<=alen;i++)
        {
            for(int j=1;j<=blen;j++)
            {
                if(a[i]==b[j])
                    c[i][j]=c[i-1][j-1]+1;
                else
                    c[i][j]=max(c[i][j-1],c[i-1][j]);
            }
        }
        cout<<c[alen][blen]<<"\n";
    }
    return 0;
}

来源：https://blog.csdn.net/Black_cat_look/article/details/100079592

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最长公共子序列