/* | |
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; | |
} | |