Submission #692656

#	Time	Username	Problem	Language	Result	Execution time	Memory
692656		saayan007	Visiting Singapore (NOI20_visitingsingapore)	C++17	4 / 100	194 ms	257120 KiB

VisitingSingapore

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vpi = vector<pi>;
using vpl = vector<pl>;

#define fur(i, a, b) for(ll i = a; i <= (ll) b; ++i)
#define ruf(i, a, b) for(ll i = a; i >= (ll) b; --i)
#define fr first
#define sc second
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define nl "\n"

#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template <class T> using iset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll inf = 1e10L;

ll cost(ll x, ll y) {
    if(y & 1) {
        return  1;
    }
    return 0;
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    ll k, n, m, a, b;
    cin >> k >> n >> m >> a >> b;
    ll v[k + 1], s[n + 1], t[m + 1];

    fur(i, 1, k) {
        cin >> v[i];
    }
    fur(i, 1, n) {
        cin >> s[i];
    }
    fur(i, 1, m) {
        cin >> t[i];
    }

    ll dp[n + 1][m + 1][4];
    fur(i, 0, n) {
        fur(j, 0, m) {
            fur(x, 0, 3)
                dp[i][j][x] = -inf;
        }
    }

    dp[1][1][3] = (s[1] == t[1] ? v[s[1]] : -inf);
    dp[1][1][0] = a;
    fur(j, 2, m) {
        dp[1][j][3] = (s[1] == t[j] ? v[s[1]] + a : -inf);
        dp[1][j][0] = a;
    }

    fur(i, 2, n) {
        dp[i][1][3] = (s[i] == t[1] ? v[s[i]] : -inf);
        dp[i][1][0] = a;
    }

    fur(i, 2, n) {
        fur(j, 2, m) {
            if(s[i] == t[j]) {
                fur(y, 0, 3)
                    dp[i][j][3] = max(dp[i][j][3], dp[i - 1][j - 1][y] + v[s[i]]);
            }
            dp[i][j][0] = max({dp[i][j][0], dp[i - 1][j - 1][3] + 2*a, dp[i - 1][j - 1][1] + a, dp[i - 1][j - 1][2] + a, dp[i - 1][j - 1][0]});
            dp[i][j][1] = max({dp[i][j][1], dp[i - 1][j][1], dp[i - 1][j][3] + a});
            dp[i][j][2] = max({dp[i][j][2], dp[i][j - 1][2], dp[i][j - 1][3] + a});
        }
    }

    ll res = -inf;
    fur(i, 1, n) {
        fur(x, 0, 3) {
            res = max(res, dp[i][m][x]);
        }
    }
    cout << res << nl;
}

• Subtask 1: 4 / 4
• Subtask 2: 0 / 6
• Subtask 3: 0 / 12
• Subtask 4: 0 / 7
• Subtask 5: 0 / 8
• Subtask 6: 0 / 13
• Subtask 7: 0 / 50