dpF - Lcs
28 Jan 2021 — Tags: None
Click to show code.
using namespace std;
const int NMAX = 3e3 + 11;
string s, t;
int mem[NMAX][NMAX];
string reconstruct(int i, int j)
{
if (i == 0 or j == 0)
return "";
if (s[i - 1] == t[j - 1])
return reconstruct(i - 1, j - 1) + s[i - 1];
else
{
if (mem[i - 1][j] > mem[i][j - 1])
return reconstruct(i - 1, j);
else
return reconstruct(i, j - 1);
}
}
string solve(void)
{
int n = s.size();
int m = t.size();
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= m; ++j)
{
if (s[i - 1] == t[j - 1])
mem[i][j] = mem[i - 1][j - 1] + 1;
else
mem[i][j] = max(mem[i - 1][j], mem[i][j - 1]);
}
}
return reconstruct(n, m);
}
int main(void)
{
cin >> s >> t;
cout << solve() << endl;
return 0;
}