LCS最长公共子序列

两个字符串A，B中，最长的公共部分（子序列可以不连续）

核心部分状态转移方程：

dp[i][j]={dp[i−1][j−1]+1,A[i]==B[j]max{dp[i−1][j],dp[i][j−1]},A[i]!=B[j] dp[i][j]=\left\{ \begin{aligned} dp[i-1][j-1]+1,A[i]==B[j] \\ max\{dp[i-1][j],dp[i][j-1]\},A[i]!=B[j]\\ \end{aligned} \right. dp[i][j]={dp[i−1][j−1]+1,A[i]==B[j]max{dp[i−1][j],dp[i][j−1]},A[i]!=B[j]​
附上代码

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

const int MAXN = 1005;

int DP[MAXN][MAXN];

int main()
{
	string a;
	string b;
	while (cin >> a >> b)
	{
		int l1 = a.size();
		int l2 = b.size();
		memset(DP, 0, sizeof(DP));
		for (int i = 1; i <= l1; i++)
			for (int j = 1; j <= l2; j++)
				if (a[i - 1] == b[j - 1])
					DP[i][j] = DP[i - 1][j - 1] + 1;
				else
					DP[i][j] = max(DP[i][j - 1], DP[i - 1][j]);
		printf("%d\n", DP[l1][l2]);
	}
	return 0;
}

继续待更