HDU 1159. Common Subsequence (DP longest common subsequence) super detailed solution

1. Title

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.
Input

abcfbc abfcab
programming contest 
abcd mnp

Output

4
2
0

2. Questions

Find the longest common subsequence length of two character strings.

3. Ideas

Dynamic programming
1. Let the array dp [i] [j] represent the longest common subsequence length before the i-th position of the string a and the j-th position of the string b. (Excluding the i-bit and the j-bit):
(1) If a [i-1] = b [j-1], the length of the longest common subsequence of the two strings increases by one bit, that is dp [i] [j] = dp [i-1] [j-1] +1.
(2) If a [i-1]! = b [j-1], then dp [] i [j] inherits The larger of dp [i-1] [j] and dp [i] [j-1], that is, dp [i] [j] = max (dp [i-1] [j], dp [i] [j-1]).
2. The boundary dp [0 ... len_a] [0] = 0, dp [0] [0 ... len_b] = 0.
3. In this way, dp [i] [j] is only related to the previous state, and the entire dp array can be obtained from the boundary. Dp [len_a] [len_b] is the answer.

4. Code

#define _CRT_SECURE_NO_WARNINGS
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 1000;
int dp[maxn][maxn];
char a[maxn],b[maxn];
int main()
{
	while (scanf("%s%s", a, b)!=EOF)
	{
		memset(dp, 0, sizeof(dp));
		int len_a = strlen(a);
		int len_b = strlen(b);
		for (int i = 0; i <= len_a; i++) 
			dp[i][0] = 0;
		for (int i = 0; i <=len_b; i++)
			dp[0][i] = 0;
		for (int i = 1 ; i <=len_a; i++)
		{
			for (int j = 1; j <= len_b; j++)
			{
				if (a[i-1] == b[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[len_a][len_b]);
	}
	return 0;
}