제출 #1126936

#제출 시각아이디문제언어결과실행 시간메모리
1126936fzyzzz_zMarathon Race 2 (JOI24_ho_t3)C++20
7 / 100
3 ms4424 KiB
#include <bits/stdc++.h> using namespace std; using ll = long long; const int N = 5e5 + 10; const int A = 1001; const ll inf = (1LL << 61); inline void amin(ll & x, ll y) { if (y < x) x = y; } #define int ll int f[N]; ll dp[2][A][A]; ll cost[2][A]; int32_t main() { ios_base::sync_with_stdio(false); cin.tie(0); int N, L; cin >> N >> L; vector<int> X(N); for (auto & x: X) cin >> x; sort(X.begin(), X.end()); vector<pair<int, int>> a(1, {X[0], 0}); for (auto x: X) { if (x == a.back().first) a.back().second++; else a.emplace_back(x, 1); } int q; cin >> q; int n = a.size(); if (n > 1000) { while (q--) { cout << "No\n"; } exit(0); } for (int i = 0; i <= L + 1; ++i) f[i] = 0; for (auto [x, c]: a) f[x + 1] += c; for (int i = 0; i <= L; ++i) f[i + 1] += f[i]; for (int t = 0; t < 2; ++t) { for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) dp[0][i][j] = dp[1][i][j] = inf; dp[t][0][n - 1] = 0; // accounted for all outside of (i, j), at 0L or 1R for (int len = n - 1; len > 0; --len) { for (int l = 0, r = len; r < n; l++, r++) { ll has = 0; ll pl = a[l].first; ll pl1 = a[l + 1].first; ll pr1 = a[r - 1].first; ll pr = a[r].first; has += f[pl]; has += f[L + 1] - f[pr + 1]; has++; amin(dp[0][l + 1][r], dp[0][l][r] + (has + a[l].second) * (pl1 - pl)); amin(dp[1][l][r - 1], dp[0][l][r] + (has) * (2LL * pr - pl - pl1) + 1LL * a[r].second * (pr - pr1)); amin(dp[1][l][r - 1], dp[1][l][r] + (has + a[r].second) * (pr - pr1)); amin(dp[0][l + 1][r], dp[1][l][r] + (has) * (pr + pr1 - 2LL * pl) + 1LL * a[l].second * (pl1 - pl)); } } for (int i = 0; i < n; ++i) { cost[t][i] = dp[t][i][i]; } } while (q--) { int st, en; ll t; cin >> st >> en >> t; t -= f[L + 1]; int lo = 0, hi = n - 1; while (lo < hi) { int md = (lo + hi) / 2; if (a[md].first >= en) hi = md; else lo = md + 1; } ll ans = inf; for (int i = max(0LL, lo - 3); i <= min(n - 1, lo + 3); ++i) { amin(ans, cost[0][i] + abs(st - a[0].first) + (ll) (f[L + 1] + 1) * (abs(en - a[n - 1].first))); amin(ans, cost[1][i] + abs(st - a[n - 1].first) + (ll) (f[L + 1] + 1) * (abs(en - a[0].first))); } if (ans <= t) { cout << "Yes\n"; } else { cout << "No\n"; } } }
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