很简单的dp题

#include <bits/stdc++.h>

using namespace std;

const int N = 60;

int T;
int dp[N][N];
char str[N][N];

int main()
{
    cin >> T;

    while (T -- )
    {
        int n;
        cin >> n;

        for (int i = 0; i < n; i ++ )
            cin >> str[i] + 1;

        int res = 0;
        memset(dp, 0, sizeof dp);
        for (int i = 0; i < n; i ++ )
        {
            for (int j = i + 1; j < n; j ++ )
            {
//              cout << i << ' ' << j << ' ';
                memset(dp, 0, sizeof dp);
                int lena = strlen(str[i] + 1), lenb = strlen(str[j] + 1);

                for (int a = 1; a <= lena; a ++ )
                    for (int b = 1; b <= lenb; b ++ )
                    {
                        if (str[i][a] == str[j][b]) dp[a][b] = max(dp[a][b], dp[a - 1][b - 1] + 1);
                        res = max(res, dp[a][b]);
                    }   

//              cout << dp[lena][lenb] << endl; 

            }
        }

        cout << res << endl;
    }
}