結果
問題 | No.777 再帰的ケーキ |
ユーザー | kakira9618 |
提出日時 | 2018-09-21 12:59:23 |
言語 | C++11 (gcc 11.4.0) |
結果 |
AC
|
実行時間 | 331 ms / 2,000 ms |
コード長 | 4,570 bytes |
コンパイル時間 | 1,984 ms |
コンパイル使用メモリ | 187,660 KB |
実行使用メモリ | 20,412 KB |
最終ジャッジ日時 | 2024-04-29 21:21:29 |
合計ジャッジ時間 | 5,414 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | AC | 1 ms
5,376 KB |
testcase_03 | AC | 2 ms
5,376 KB |
testcase_04 | AC | 2 ms
5,376 KB |
testcase_05 | AC | 2 ms
5,376 KB |
testcase_06 | AC | 7 ms
5,376 KB |
testcase_07 | AC | 14 ms
5,376 KB |
testcase_08 | AC | 1 ms
5,376 KB |
testcase_09 | AC | 1 ms
5,376 KB |
testcase_10 | AC | 1 ms
5,376 KB |
testcase_11 | AC | 2 ms
5,376 KB |
testcase_12 | AC | 2 ms
5,376 KB |
testcase_13 | AC | 2 ms
5,376 KB |
testcase_14 | AC | 2 ms
5,376 KB |
testcase_15 | AC | 2 ms
5,376 KB |
testcase_16 | AC | 2 ms
5,376 KB |
testcase_17 | AC | 2 ms
5,376 KB |
testcase_18 | AC | 8 ms
5,376 KB |
testcase_19 | AC | 14 ms
5,376 KB |
testcase_20 | AC | 15 ms
5,376 KB |
testcase_21 | AC | 15 ms
5,376 KB |
testcase_22 | AC | 15 ms
5,376 KB |
testcase_23 | AC | 2 ms
5,376 KB |
testcase_24 | AC | 2 ms
5,376 KB |
testcase_25 | AC | 8 ms
5,376 KB |
testcase_26 | AC | 9 ms
5,376 KB |
testcase_27 | AC | 2 ms
5,376 KB |
testcase_28 | AC | 326 ms
20,280 KB |
testcase_29 | AC | 331 ms
20,412 KB |
testcase_30 | AC | 320 ms
20,284 KB |
testcase_31 | AC | 324 ms
20,152 KB |
testcase_32 | AC | 194 ms
20,284 KB |
testcase_33 | AC | 35 ms
5,376 KB |
testcase_34 | AC | 52 ms
5,376 KB |
testcase_35 | AC | 198 ms
11,728 KB |
testcase_36 | AC | 197 ms
20,284 KB |
ソースコード
#include <bits/stdc++.h> using namespace std; #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define rep(i,n) for(int i=0;i<(n);i++) constexpr int MOD = 1000000007; typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii; constexpr int dx[] = {1, 0, -1, 0, 1, 1, -1, -1}; constexpr int dy[] = {0, -1, 0, 1, 1, -1, -1, 1}; template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){os << "["; for (const auto &v : vec) {os << v << ","; } os << "]"; return os; } template <typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &p){os << "(" << p.first << ", " << p.second << ")"; return os;} struct Phone { int h, w, t; bool operator<(const Phone &rhs) const { if (h == rhs.h) { return w > rhs.w; } return h < rhs.h; } }; struct BIT { vector<ll> data; BIT(int n): data(n + 1) {} ll getmax(int a) { ll ret = 0; for(int i = a; i > 0; i -= i & -i) { ret = max(ret, data[i]); } return ret; } void update(int a, ll w) { for(int i = a; i < data.size(); i += i & -i) { data[i] = max(data[i], w); } } }; // BIT解(想定解) ll solve1(int N, vector<Phone> &P) { sort(all(P)); vector<int> W(N); for (int i = 0; i < N; i++) { W[i] = P[i].w; } sort(all(W)); W.erase(unique(W.begin(), W.end()), W.end()); BIT bit(N + 2); for(int i = 0; i < N; i++) { int k = lower_bound(all(W), P[i].w) - W.begin() - 1; k += 2; // for BIT ll mi = bit.getmax(k); bit.update(k + 1, mi + P[i].t); } return bit.getmax(N + 2); } // Naive DP解(O(N^2), TLE) ll solve2(int N, vector<Phone> &P) { sort(all(P)); vector<ll> dp(N, 0); for(int i = 0; i < N; i++) { ll ma = 0; for(int j = 0; j < i; j++) { if (P[j].w >= P[i].w) continue; ma = max(ma, dp[j]); } dp[i] = ma + P[i].t; } return *max_element(all(dp)); } // 二分探索 LIS解(O(N log N), T_i = 1) ll LIS(vector<ll> &A) { vector<ll> dp; for(int i = 0; i < A.size(); i++) { int k = lower_bound(all(dp), A[i]) - dp.begin(); if (k == dp.size()) dp.push_back(A[i]); else dp[k] = A[i]; } return dp.size(); } ll LIS_dp(vector<ll> &A) { vector<ll> dp(A.size()); for (int i = 0; i < A.size(); i++) { ll ma = 0; for (int j = 0; j < i; j++) { if (A[j] >= A[i]) continue; ma = max(ma, dp[j]); } dp[i] = ma + 1; } return *max_element(all(dp)); } ll solve3(int N, vector<Phone> &P) { sort(all(P)); vector<ll> W(N); for (int i = 0; i < N; i++) { W[i] = P[i].w; } return LIS(W); } ll solve4(int N, vector<Phone> &P) { sort(all(P)); vector<ll> W(N); for (int i = 0; i < N; i++) { W[i] = P[i].w; } return LIS_dp(W); } ll solve5(int N, vector<Phone> &P) { sort(all(P)); ll ans = 0; for(int i = 0; i < 1 << N; i++) { vector<ll> W; ll sum = 0; for(int j = 0; j < N; j++) { if (i >> j & 1) { W.push_back(P[j].w); sum += P[j].t; } } bool flag = true; for(int j = 0; j + 1 < W.size(); j++) { if (W[j] >= W[j + 1]) { flag = false; break; } } if (!flag) continue; ans = max(ans, sum); } return ans; } void solve() { int N; cin >> N; if (N < 1 || N > 200000) exit(1); map<pii, int> M; for(int i = 0; i < N; i++) { int h, w, t; cin >> h >> w >> t; if (h < 1 || h > 1000000000) exit(1); if (w < 1 || w > 1000000000) exit(1); if (t < 1 || t > 1000000000) exit(1); M[{h, w}] = max(M[{h, w}], t); } vector<Phone> P; for(auto &&x : M) { P.push_back({x.first.first, x.first.second, x.second}); } N = P.size(); bool solve1_f = true; bool solve2_f = N <= 2000; bool solve3_f = true; for (int i = 0; i < N; i++) { if (P[i].t != 1) solve3_f = false; } bool solve4_f = solve3_f && (N <= 2000); bool solve5_f = N <= 16; vector<ll> ans; if (solve1_f) ans.push_back(solve1(N, P)); if (solve2_f) ans.push_back(solve2(N, P)); if (solve3_f) ans.push_back(solve3(N, P)); if (solve4_f) ans.push_back(solve4(N, P)); if (solve5_f) ans.push_back(solve5(N, P)); set<ll> S; for (int i = 0; i < ans.size(); i++) { S.insert(ans[i]); } if (S.size() > 1) { cout << "Error!" << endl; for (int i = 0; i < ans.size(); i++) { cout << ans[i] << endl; } } else { cout << ans[0] << endl; } } int main() { std::cin.tie(0); std::ios::sync_with_stdio(false); cout.setf(ios::fixed); cout.precision(16); solve(); return 0; }