제출 #626149

#제출 시각아이디문제언어결과실행 시간메모리
626149I_love_Hoang_YenCatfish Farm (IOI22_fish)C++17
100 / 100
831 ms360284 KiB
#include "bits/stdc++.h" using namespace std; #define int long long #define i_1 jakcjacjl struct Fish { int col, row; int weight; }; bool operator < (const Fish& a, const Fish& b) { if (a.col != b.col) return a.col < b.col; return a.row < b.row; } void upMax(int& f, int val) { if (val > f) f = val; } // sub1 - 3 {{{ // fishes are on even columns -> build piers on odd columns // & catch all fishes int sub1(const std::vector<Fish>& fishes) { int res = 0; for (const auto& fish : fishes) { res += fish.weight; } return res; } // fishes are on first 2 columns int sub2(int n, const std::vector<Fish>& fishes) { std::vector<int> zeroes(n); // prefix sum of fish weights at column == 0 std::vector<int> ones(n); // prefix sum of fish weights at column == 1 for (const auto& fish : fishes) { if (fish.col == 0) zeroes[fish.row] += fish.weight; if (fish.col == 1) ones[fish.row] += fish.weight; } std::partial_sum(zeroes.begin(), zeroes.end(), zeroes.begin()); std::partial_sum(ones.begin(), ones.end(), ones.begin()); int res = ones.back(); // init: only catch fishes at column == 1 for (int i = 0; i < n; ++i) { // build pier until at column 1, row 0-i if (n == 2) upMax(res, zeroes[i]); else upMax(res, zeroes[i] + ones.back() - ones[i]); } return res; } // all fishes are on row == 0 int sub3(int n, const std::vector<Fish>& fishes) { std::vector<int> weights(n); // weights[i] = weight of fish at column i for (const auto& fish : fishes) { weights[fish.col] += fish.weight; } // f[i] = best strategy if we BUILD PIER AT i, only considering col 0..i // i-4 i-3 i-2 i-1 i std::vector<int> f(n); f[0] = 0; for (int i = 1; i < n; ++i) { f[i] = std::max(f[i-1], weights[i-1]); if (i >= 2) { upMax(f[i], f[i-2] + weights[i-1]); } if (i >= 3) { upMax(f[i], f[i-3] + weights[i-2] + weights[i-1]); } } int res = 0; for (int i = 0; i < n; ++i) { int cur = f[i]; if (i + 1 < n) cur += weights[i+1]; upMax(res, cur); } return res; } // }}} // sub 5 N <= 300 {{{ int sub5(int n, const std::vector<Fish>& fishes) { // Init weights[i][j] = sum of fish on column i, from row 0 -> row j std::vector<std::vector<int>> weights(n, std::vector<int> (n, 0)); for (const auto& fish : fishes) { weights[fish.col][fish.row] += fish.weight; } for (int col = 0; col < n; ++col) { std::partial_sum(weights[col].begin(), weights[col].end(), weights[col].begin()); } // f[c][r] = best strategy if we last BUILD PIER AT column c, row r // only considering fishes <= (c, r) // g[c][r] = similar to f[c][r] but consider fishes at column c, in row [r, n-1] std::vector<std::vector<int>> f(n, std::vector<int> (n, 0)), g(n, std::vector<int> (n, 0)); // f <= g for (int c = 1; c < n; ++c) { for (int r = 0; r < n; ++r) { // this is first pier f[c][r] = g[c][r] = weights[c-1][r]; // last pier at column i-1 for (int lastRow = 0; lastRow < n; ++lastRow) { if (lastRow <= r) { int cur = std::max( f[c-1][lastRow] + weights[c-1][r] - weights[c-1][lastRow], g[c-1][lastRow]); upMax(f[c][r], cur); upMax(g[c][r], cur); } else { upMax(f[c][r], g[c-1][lastRow]); upMax(g[c][r], g[c-1][lastRow] + weights[c][lastRow] - weights[c][r]); } } // last pier at column i-2 if (c >= 2) { for (int lastRow = 0; lastRow < n; ++lastRow) { int cur = g[c-2][lastRow] + weights[c-1][std::max(lastRow, r)]; upMax(f[c][r], cur); upMax(g[c][r], cur); } } // last pier at column i-3 if (c >= 3) { for (int lastRow = 0; lastRow < n; ++lastRow) { int cur = g[c-3][lastRow] + weights[c-2][lastRow] + weights[c-1][r]; upMax(f[c][r], cur); upMax(g[c][r], cur); } } } } int res = 0; for (int c = 0; c < n; ++c) { for (int r = 0; r < n; ++r) { assert(g[c][r] >= f[c][r]); int cur = g[c][r]; if (c + 1 < n) { cur += weights[c+1][r]; } upMax(res, cur); } } return res; } // }}} // RMQ {{{ // // Sparse table // Usage: // RMQ<int, _min> st(v); // // Note: // - doesn't work for empty range // // Tested: // - https://judge.yosupo.jp/problem/staticrmq template<class T, T (*op) (T, T)> struct RMQ { RMQ() = default; RMQ(const vector<T>& v) : t{v}, n{(int) v.size()} { for (int k = 1; (1<<k) <= n; ++k) { t.emplace_back(n - (1<<k) + 1); for (int i = 0; i + (1<<k) <= n; ++i) { t[k][i] = op(t[k-1][i], t[k-1][i + (1<<(k-1))]); } } } // get range [l, r-1] // doesn't work for empty range T get(int l, int r) const { if (l == n) return 0; assert(0 <= l && l < r && r <= n); int k = __lg(r - l); return op(t[k][l], t[k][r - (1<<k)]); } T get_from(int l) const { return get(l, n); } T get_all() const { return get(0, n); } private: vector<vector<T>> t; int n; }; template<class T> T _min(T a, T b) { return b < a ? b : a; } template<class T> T _max(T a, T b) { return a < b ? b : a; } // }}} // N <= 3000 {{{ int sub6(int n, const std::vector<Fish>& fishes) { // Init weights[i][j] = sum of fish on column i, from row 0 -> row j std::vector<std::vector<int>> weights(n, std::vector<int> (n, 0)); for (const auto& fish : fishes) { weights[fish.col][fish.row] += fish.weight; } for (int col = 0; col < n; ++col) { std::partial_sum(weights[col].begin(), weights[col].end(), weights[col].begin()); } // f[c][r] = best strategy if we last BUILD PIER AT column c, row r // only considering fishes <= (c, r) // g[c][r] = similar to f[c][r] but consider fishes at column c, in row [r, n-1] std::vector<std::vector<int>> f(n, std::vector<int> (n, 0)), g(n, std::vector<int> (n, 0)); std::vector<RMQ<int, _max>> rmq_g(n), rmq_g_with_next_col(n), rmq_f_with_next_col(n); for (int c = 0; c < n; ++c) { // compute {{{ if (c > 0) { for (int r = 0; r < n; ++r) { // this is first pier f[c][r] = weights[c-1][r]; // last pier at column i-3 if (c >= 3) { upMax(f[c][r], rmq_g_with_next_col[c-3].get_all() + weights[c-1][r]); } // last pier at column i-2 if (c >= 2) { upMax(f[c][r], std::max( rmq_g[c-2].get_all() + weights[c-1][r], rmq_g_with_next_col[c-2].get_all())); } g[c][r] = f[c][r]; // last pier at column i-1 if (c >= 1) { // last row <= r int cur = std::max( rmq_g[c-1].get(0, r+1), rmq_f_with_next_col[c-1].get(0, r+1) + weights[c-1][r]); upMax(f[c][r], cur); upMax(g[c][r], cur); // last row > r if (r + 1 < n) { upMax(f[c][r], rmq_g[c-1].get(r+1, n)); upMax(g[c][r], rmq_g_with_next_col[c-1].get(r+1, n) - weights[c][r]); } } } } // }}} // aggregate {{{ rmq_g[c] = RMQ<int, _max> (g[c]); if (c + 1 < n) { std::vector<int> g_with_next_col(n), f_with_next_col(n); for (int r = 0; r < n; ++r) { g_with_next_col[r] = g[c][r] + weights[c+1][r]; f_with_next_col[r] = f[c][r] - weights[c][r]; } rmq_g_with_next_col[c] = RMQ<int, _max> (g_with_next_col); rmq_f_with_next_col[c] = RMQ<int, _max> (f_with_next_col); } // }}} } int res = 0; for (int c = 0; c < n; ++c) { for (int r = 0; r < n; ++r) { assert(g[c][r] >= f[c][r]); int cur = g[c][r]; if (c + 1 < n) { cur += weights[c+1][r]; } upMax(res, cur); } } return res; } // }}} // AC {{{ int sub7(int n, const std::vector<Fish>& fishes) { std::vector<std::vector<int>> rows(n); // rows[c] = important coordinates at col c std::vector<std::vector<int>> weights(n); // prefix sum of weights std::vector<std::vector<std::pair<int,int>>> fishesAt(n); // stores {row, weight} for (const auto& fish : fishes) { int c = fish.col; rows[c].push_back(fish.row); if (c > 0) rows[c-1].push_back(fish.row); if (c + 1 < n) rows[c+1].push_back(fish.row); fishesAt[c].push_back({fish.row, fish.weight}); } for (int c = 0; c < n; ++c) { rows[c].push_back(-1); std::sort(rows[c].begin(), rows[c].end()); rows[c].erase(std::unique(rows[c].begin(), rows[c].end()), rows[c].end()); std::sort(fishesAt[c].begin(), fishesAt[c].end()); weights[c].resize(rows[c].size()); int fish_id = 0; for (int i = 0; i < (int) rows[c].size(); ++i) { if (i > 0) weights[c][i] = weights[c][i-1]; while (fish_id < (int) fishesAt[c].size() && fishesAt[c][fish_id].first <= rows[c][i]) { weights[c][i] += fishesAt[c][fish_id].second; ++fish_id; } } } // f[c][r] = best strategy if we last BUILD PIER AT column c, row r // only considering fishes <= (c, r) // g[c][r] = similar to f[c][r] but consider fishes at column c, in row [r, n-1] std::vector<std::vector<int>> f(n), g(n); std::vector<RMQ<int, _max>> rmq_g(n), rmq_g_with_next_col(n), rmq_f_with_next_col(n); for (int c = 0; c < n; ++c) { int sz = static_cast<int> (rows[c].size()); f[c] = g[c] = std::vector<int> (sz, 0); // compute {{{ if (c > 0) { for (int i = 0; i < sz; ++i) { int i_1 = std::upper_bound(rows[c-1].begin(), rows[c-1].end(), rows[c][i]) - rows[c-1].begin() - 1; // this is first pier f[c][i] = weights[c-1][i_1]; // last pier at column i-3 if (c >= 3) { upMax(f[c][i], rmq_g_with_next_col[c-3].get_all() + weights[c-1][i_1]); } // last pier at column i-2 if (c >= 2) { upMax(f[c][i], std::max( rmq_g[c-2].get_all() + weights[c-1][i_1], rmq_g_with_next_col[c-2].get_all())); } g[c][i] = f[c][i]; // last pier at column i-1 if (c >= 1) { // last row <= r int cur = std::max( rmq_g[c-1].get(0, i_1+1), rmq_f_with_next_col[c-1].get(0, i_1+1) + weights[c-1][i_1]); upMax(f[c][i], cur); upMax(g[c][i], cur); // last row > r upMax(f[c][i], rmq_g[c-1].get_from(i_1+1)); upMax(g[c][i], rmq_g_with_next_col[c-1].get_from(i_1+1) - weights[c][i]); } } } // }}} // aggregate {{{ rmq_g[c] = RMQ<int, _max> (g[c]); if (c + 1 < n) { std::vector<int> g_with_next_col(sz), f_with_next_col(sz); for (int i = 0; i < sz; ++i) { int i_1 = std::upper_bound(rows[c+1].begin(), rows[c+1].end(), rows[c][i]) - rows[c+1].begin() - 1; g_with_next_col[i] = g[c][i] + weights[c+1][i_1]; f_with_next_col[i] = f[c][i] - weights[c][i]; } rmq_g_with_next_col[c] = RMQ<int, _max> (g_with_next_col); rmq_f_with_next_col[c] = RMQ<int, _max> (f_with_next_col); } // }}} } int res = 0; for (int c = 0; c < n; ++c) { int sz = rows[c].size(); for (int i = 0; i < sz; ++i) { int cur = g[c][i]; if (c + 1 < n) { int i_1 = std::upper_bound(rows[c+1].begin(), rows[c+1].end(), rows[c][i]) - rows[c+1].begin() - 1; cur += weights[c+1][i_1]; } upMax(res, cur); } } return res; } // }}} #undef int long long max_weights( int n, int nFish, std::vector<int> xs, std::vector<int> ys, std::vector<int> ws) { std::vector<Fish> fishes; for (int i = 0; i < nFish; ++i) { fishes.push_back({xs[i], ys[i], ws[i]}); } if (std::all_of(xs.begin(), xs.end(), [] (int x) { return x % 2 == 0; })) { return sub1(fishes); } if (*std::max_element(xs.begin(), xs.end()) <= 1) { return sub2(n, fishes); } if (*std::max_element(ys.begin(), ys.end()) == 0) { return sub3(n, fishes); } return sub7(n, fishes); }
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