Submission #693005

#	Time	Username	Problem	Language	Result	Execution time	Memory
693005		saayan007	Visiting Singapore (NOI20_visitingsingapore)	C++17	100 / 100	654 ms	848 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"

const ll inf = 1e10L;

ll k, n, m, a, b;
ll cost(ll y, ll x) {
    ll l, r;

    ll ans = 0;

    l = y%2, r = x%2;
    if(r == 0) {
        ans -= b;
        if(l == 1)
            ans -= a;
    }

    y /= 2;
    x /= 2;

    l = y%2, r = x%2;
    if(r == 0) {
        ans -= b;
        if(l == 1)
            ans -= a;
    }
    return ans;
}


int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin >> k >> n >> m >> a >> b;
    a = -a;
    b = -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[2][m + 1][4];
    fur(i, 0, 1)
        fur(j, 0, m)
            fur(x, 0, 3)
                dp[i][j][x] = -inf;

    ll res = -a -m*b;

    fur(i, 1, n) {
        ll cur = i % 2;
        ll prv = 1 - cur;

        fur(j, 1, m) {
            if(s[i] == t[j]) {
                fur(y, 0, 3) {
                    dp[cur][j][3] = max({dp[cur][j][3], dp[prv][j - 1][y] + v[t[j]] + cost(y, 3), v[t[j]] + (j == 1 ? 0 : - a - (j - 1) * b)});
                    res = max(res, dp[cur][j][3] + (j == m ? 0 : - a - b * (m - j)));
                }
            }

            dp[cur][j][0] = max({dp[cur][j][0], dp[prv][j][0] + cost(0, 0), dp[prv][j][2] + cost(2, 0), dp[cur][j - 1][0] + cost(0, 0), dp[cur][j - 1][1] + cost(1, 0)});
            fur(y, 0, 3) {
                dp[cur][j][0] = max(dp[cur][j][0], dp[prv][j - 1][y] + cost(y, 0));
            }
            dp[cur][j][1] = max({dp[cur][j][1], dp[prv][j][1] + cost(1, 1), dp[prv][j][3] + cost(3, 1)});
            dp[cur][j][2] = max({dp[cur][j][2], dp[cur][j - 1][2] + cost(2, 2), dp[cur][j - 1][3] + cost(3, 2)});
        }

        fur(j, 0, m) {
            fur(x, 0, 3) {
                dp[prv][j][x] = -inf;
            }
        }
    }
    cout << res << nl;
}

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