結果
問題 |
No.901 K-ary εxtrεεmε
|
ユーザー |
|
提出日時 | 2024-09-04 17:21:13 |
言語 | C++23 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 362 ms / 3,000 ms |
コード長 | 5,030 bytes |
コンパイル時間 | 3,623 ms |
コンパイル使用メモリ | 264,504 KB |
実行使用メモリ | 29,048 KB |
最終ジャッジ日時 | 2024-09-04 17:21:26 |
合計ジャッジ時間 | 13,337 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 1 |
other | AC * 29 |
ソースコード
#include <bits/stdc++.h> using namespace std; template <typename T> struct graph { struct edge { int from; int to; T cost; }; vector<edge> edges; vector<vector<int>> g; int n; graph(int _n) : n(_n) { g.resize(n); } virtual int add(int from, int to, T cost) = 0; }; template <typename T> struct forest : graph<T> { using graph<T>::edges; using graph<T>::g; using graph<T>::n; forest(int _n) : graph<T>(_n) {} int add(int from, int to, T cost = 1) { assert(0 <= from && from < n && 0 <= to && to < n); int id = (int)edges.size(); assert(id < n - 1); g[from].push_back(id); g[to].push_back(id); edges.push_back({from, to, cost}); return id; } }; template <typename T> struct dfs_forest : forest<T> { using forest<T>::edges; using forest<T>::g; using forest<T>::n; vector<int> pv; vector<int> pe; vector<int> order; vector<int> pos; vector<int> end; vector<int> sz; vector<int> root; vector<int> depth; vector<T> dist; dfs_forest(int _n) : forest<T>(_n) {} void init() { pv = vector<int>(n, -1); pe = vector<int>(n, -1); order.clear(); pos = vector<int>(n, -1); end = vector<int>(n, -1); sz = vector<int>(n, 0); root = vector<int>(n, -1); depth = vector<int>(n, -1); dist = vector<T>(n); } void clear() { pv.clear(); pe.clear(); order.clear(); pos.clear(); end.clear(); sz.clear(); root.clear(); depth.clear(); dist.clear(); } void dfs(int v, bool clear_order = false) { if(pv.empty()) init(); else if(clear_order) order.clear(); do_dfs_from(v); } void dfs_all() { init(); for(int v = 0; v < n; v++) { if(depth[v] == -1) do_dfs_from(v); } assert((int)order.size() == n); } private: void do_dfs(int v) { pos[v] = (int) order.size(); order.push_back(v); sz[v] = 1; for (int id : g[v]) { if(id == pe[v]) continue; auto &e = edges[id]; int to = e.from ^ e.to ^ v; depth[to] = depth[v] + 1; dist[to] = dist[v] + e.cost; pv[to] = v; pe[to] = id; root[to] = (root[v] != -1 ? root[v] : to); do_dfs(to); sz[v] += sz[to]; } end[v] = (int) order.size() - 1; } void do_dfs_from(int v) { depth[v] = 0; dist[v] = T{}; root[v] = v; pv[v] = pe[v] = -1; do_dfs(v); } }; template <typename T> struct lca_forest : dfs_forest<T> { using dfs_forest<T>::edges; using dfs_forest<T>::g; using dfs_forest<T>::n; using dfs_forest<T>::pv; using dfs_forest<T>::pos; using dfs_forest<T>::end; using dfs_forest<T>::depth; int h; vector<vector<int>> pr; lca_forest(int _n) : dfs_forest<T>(_n) {} inline void build_lca() { assert(!pv.empty()); int max_depth = 0; for(int i = 0; i < n; i++) max_depth = max(max_depth, depth[i]); h = 1; while((1 << h) <= max_depth) h++; pr.resize(n); for(int i = 0; i < n; i++) { pr[i].resize(h); pr[i][0] = pv[i]; } for(int j = 1; j < h; j++) { for(int i = 0; i < n; i++) { pr[i][j] = (pr[i][j - 1] == -1 ? -1 : pr[pr[i][j - 1]][j - 1]); } } } inline bool anc(int x, int y) { return (pos[x] <= pos[y] && end[y] <= end[x]); } inline int go_up(int x, int up) { assert(!pr.empty()); up = min(up, (1 << h) - 1); for(int j = h - 1; j >= 0; j--) { if(up & (1 << j)) { x = pr[x][j]; if(x == -1) break; } } return x; } inline int lca(int x, int y) { assert(!pr.empty()); if(anc(x, y)) return x; if(anc(y, x)) return y; for(int j = h - 1; j >= 0; j--) { if(pr[x][j] != -1 && !anc(pr[x][j], y)) x = pr[x][j]; } return pr[x][0]; } }; int main() { int n; cin >> n; lca_forest<long long> g(n); for(int i = 1; i < n; i++) { int u, v, w; cin >> u >> v >> w; g.add(u, v, w); } g.dfs_all(); g.build_lca(); int q; cin >> q; while(q--) { int K; cin >> K; vector<int> vs; for(int i = 0; i < K; i++) { int u; cin >> u; vs.push_back(u); } sort(begin(vs), end(vs), [&](int u, int v) -> bool { return g.pos[u] < g.pos[v]; }); vs.push_back(vs[0]); long long ans = 0; for(int i = 0; i < K; i++) { ans += g.dist[vs[i]] + g.dist[vs[i + 1]] - 2 * g.dist[g.lca(vs[i], vs[i + 1])]; } cout << ans/2 << endl; } }