| /* | |
| dynamic programming > longest common subsequence (LCS) | |
| difficulty: easy | |
| date: 29/Apr/2020 | |
| by: @brpapa | |
| */ | |
| using namespace std; | |
| vector<int> p; | |
| string a; int N; | |
| string b; int M; | |
| int memo[2020][2020]; | |
| int dp(int i, int j) { | |
| // atual a[i] e b[j] | |
| if (i == N || j == M) return 0; | |
| int &ans = memo[i][j]; | |
| if (ans != -1) return ans; | |
| if (a[i] == b[j]) | |
| return ans = p[a[i]-'a'] + dp(i+1, j+1); | |
| return ans = max(dp(i+1, j), dp(i, j+1)); | |
| } | |
| int main() { | |
| cin >> N >> M; | |
| p.resize(26); for (int &price: p) cin >> price; | |
| cin >> a >> b; | |
| memset(memo, -1, sizeof memo); | |
| cout << dp(0, 0) << endl; | |
| return 0; | |
| } | |