【ACWing】897. The longest common subsequence

Subject address:

https://www.acwing.com/problem/content/899/

Given two lengths are $N$ and$M$ 's string$A$ and$B$ , seeking to be bothAAThe subsequence of$A$ isBB againWhat is the longest string length of the subsequence of$B$ ?

Data range:
$1\le N,M\le 1000$

The idea is dynamic programming. Let $f\left[i\right]\left[j\right]$ Yes$A$ basis before$i$ characters and$B$ 's formerjjThe length of the longest common subsequence of$j$ characters. When$A[i]\ne When\; B\left[j\right]$ , obviouslyf [i] [j] = max ⁡ {f [i − 1] [j], f [i] [j − 1]} f[i][j]=\max\ {f[i-1][j],f[i][j-1]\}f[i][j]=max{ f[i−1][j],f[i][j−1]}；当$A\left[i\right]=B\left[j\right]$ target time$f\left[i\right]\left[j\right]=1+f[i-1][j-1]$ , the reason is, first of all it is obvious that$f[i][j]\ge 1+f[i-1][j-1]$ and secondly, if$f\left[i\right]\left[j\right]>1+f[i-1][j-1]$ , if$f\left[i\right][j]\; The$ last equivalent character is$A\left[i\right]$ sum$B\left[j\right]$， 那么$f\left[i\right]\left[j\right]=1+f[i-1][j-1]$ , contradiction; if not, thenf [i] [j] = max ⁡ {f [i − 1] [j], f [i] [j − 1]} f[i][j]=\ max\{f[i-1][j],f[i][j-1]\}f[i][j]=max{ f[i−1][j],f[i][j−1]}，如果是$f[i-1][j]>1+f[i-1][j-1]$ , also consider$f[i-1]$ The last character of$[$$j$$]$ , if$t[j]$就会得出$1+f[i-1][j-1]>1+f[i-1][j-1]$ contradiction; if it is not used,$f[i-1][j-1]>1+f[i-1][j-1]$ Also contradictory. So$f[i][j]=1+f[i-1][j-1]$ . code show as below:

#include <iostream>
using namespace std;

const int N = 1010;
string s, t;
int n, m;
int f[N][N];

int main() {
    
    
    cin >> n >> m;
    cin >> s >> t;

    s = ' ' + s;
    t = ' ' + t;

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) 
            if (s[i] == t[j]) f[i][j] = 1 + f[i - 1][j - 1];
            else f[i][j] = max(f[i - 1][j], f[i][j - 1]);

    cout << f[n][m] << endl;

    return 0;
}

Time and space complexity $O(NM)$。