제출 #608510

#제출 시각아이디문제언어결과실행 시간메모리
608510dxz05Plahte (COCI17_plahte)C++14
160 / 160
312 ms29204 KiB
#pragma GCC optimize("Ofast,O2,O3,unroll-loops") #pragma GCC target("avx2") #include <bits/stdc++.h> using namespace std; int main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); int n, m; cin >> n >> m; vector<int> x1(n + m), x2(n + m), y1(n + m), y2(n + m); vector<pair<int, int>> events; vector<int> color(n + m); for (int i = 0; i < n + m; i++){ cin >> x1[i] >> y1[i]; if (i < n){ cin >> x2[i] >> y2[i]; color[i] = 0; } else { x2[i] = x1[i], y2[i] = y1[i]; cin >> color[i]; } events.emplace_back(i, 1); events.emplace_back(i, 2); } function<bool(int, int)> contains = [&](int i, int j){ return x1[i] <= x1[j] && x2[j] <= x2[i] && y1[i] <= y1[j] && y2[j] <= y2[i]; }; sort(events.begin(), events.end(), [&](pair<int, int> foo, pair<int, int> hoo){ int i = foo.first, j = hoo.first; int ti = foo.second, tj = hoo.second; int xi = (ti == 1 ? x1[i] : x2[i]), xj = (tj == 1 ? x1[j] : x2[j]); return make_tuple(xi, ti, i >= n) < make_tuple(xj, tj, j >= n); }); set<pair<int, int>> s; vector<int> p(n + m, -1); for (auto now : events){ int i = now.first, t = now.second; if (t == 1){ auto it = s.lower_bound(make_pair(y1[i], -1)); if (it == s.end()){ p[i] = -1; } else { int j = it->second; if (contains(j, i)){ p[i] = j; } else { p[i] = p[j]; } } s.insert(make_pair(y1[i], i)); s.insert(make_pair(y2[i], i)); } else { s.erase(make_pair(y1[i], i)); s.erase(make_pair(y2[i], i)); } } vector<vector<int>> g(n + m); for (int i = 0; i < n + m; i++){ if (p[i] != -1) g[p[i]].push_back(i); } vector<int> ans(n + m); vector<set<int>> sub(n + m); function<void(int)> dfs = [&](int v){ int big = -1; for (int u : g[v]){ dfs(u); if (big == -1 || sub[u].size() > sub[big].size()) big = u; } if (big != -1){ swap(sub[v], sub[big]); sub[big].clear(); } sub[v].insert(color[v]); for (int u : g[v]){ if (u != big){ for (int x : sub[u]){ sub[v].insert(x); } sub[u].clear(); } } ans[v] = (int)sub[v].size(); }; for (int i = 0; i < n; i++){ if (p[i] == -1) dfs(i); } for (int i = 0; i < n; i++) cout << ans[i] - 1 << '\n'; return 0; }
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