結果
問題 | No.399 動的な領主 |
ユーザー | zimpha |
提出日時 | 2019-10-27 16:33:05 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 84 ms / 2,000 ms |
コード長 | 3,922 bytes |
コンパイル時間 | 2,432 ms |
コンパイル使用メモリ | 211,012 KB |
実行使用メモリ | 12,256 KB |
最終ジャッジ日時 | 2024-09-14 21:03:53 |
合計ジャッジ時間 | 4,372 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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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
5,248 KB |
testcase_02 | AC | 2 ms
5,376 KB |
testcase_03 | AC | 2 ms
5,376 KB |
testcase_04 | AC | 2 ms
5,376 KB |
testcase_05 | AC | 8 ms
5,376 KB |
testcase_06 | AC | 84 ms
11,724 KB |
testcase_07 | AC | 82 ms
11,732 KB |
testcase_08 | AC | 80 ms
11,784 KB |
testcase_09 | AC | 80 ms
11,656 KB |
testcase_10 | AC | 3 ms
5,376 KB |
testcase_11 | AC | 8 ms
5,376 KB |
testcase_12 | AC | 65 ms
12,256 KB |
testcase_13 | AC | 64 ms
12,244 KB |
testcase_14 | AC | 56 ms
11,700 KB |
testcase_15 | AC | 56 ms
11,824 KB |
testcase_16 | AC | 57 ms
11,792 KB |
testcase_17 | AC | 81 ms
11,784 KB |
testcase_18 | AC | 77 ms
11,792 KB |
ソースコード
#include "bits/stdc++.h" using namespace std; #define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i)) #define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i)) #define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i)) static const int INF = 0x3f3f3f3f; static const long long INFL = 0x3f3f3f3f3f3f3f3fLL; typedef vector<int> vi; typedef pair<int, int> pii; typedef vector<pair<int, int> > vpii; typedef long long ll; template<typename T, typename U> static void amin(T &x, U y) { if(y < x) x = y; } template<typename T, typename U> static void amax(T &x, U y) { if(x < y) x = y; } vector<int> t_parent; vi t_ord; void tree_getorder(const vector<vi> &g, int root) { int n = g.size(); t_parent.assign(n, -1); t_ord.clear(); vector<int> stk; stk.push_back(root); while(!stk.empty()) { int i = stk.back(); stk.pop_back(); t_ord.push_back(i); for(int j = (int)g[i].size() - 1; j >= 0; j --) { int c = g[i][j]; if(t_parent[c] == -1 && c != root) stk.push_back(c); else t_parent[i] = c; } } } // Schieber-Vishkin algorithm of LCA in O(n) ~ O(1) class SchieberVishkinLCA { public: // order[]: preorder of the vertex in the tree // parents[]: direct parent of vertex // root: root of the tree void build(const std::vector<int> &order, const std::vector<int> &parents, int root) { const int n = order.size(); indices.resize(n); ascendant.resize(n); head.resize(n); for (int i = 0; i < n; ++i) indices[order[i]] = i + 1; inlabel.assign(indices.begin(), indices.end()); for (int i = n - 1; i > 0; --i) { int v = order[i], p = parents[v]; if (lowbit(inlabel[p]) < lowbit(inlabel[v])) { inlabel[p] = inlabel[v]; } } ascendant[root] = 0; for (int i = 1; i < n; ++i) { int v = order[i], p = parents[v]; ascendant[v] = ascendant[p] | lowbit(inlabel[v]); } head[0] = root; for (int i = 1; i < n; ++i) { int v = order[i], p = parents[v]; if (inlabel[v] != inlabel[p]) head[indices[v] - 1] = p; else head[indices[v] - 1] = head[indices[p] - 1]; } } // return the LCA of u and v in O(1) int query(int u, int v) const { uint Iv = inlabel[v], Iu = inlabel[u]; uint hIv = lowbit(Iv), hIu = lowbit(Iu); uint mask = highbit((Iv ^ Iu) | hIv | hIu); uint j = lowbit(ascendant[v] & ascendant[u] & ~mask); int x, y; if (j == hIv) x = v; else { mask = highbit(ascendant[v] & (j - 1)); x = head[(indices[v] & ~mask | (mask + 1)) - 1]; } if (j == hIu) y = u; else { mask = highbit(ascendant[u] & (j - 1)); y = head[(indices[u] & ~mask | (mask + 1)) - 1]; } return indices[x] < indices[y] ? x : y; } private: using uint = unsigned int; static uint lowbit(uint x) { return x & (~x + 1); } static uint highbit(uint x) { x |= x >> 1; x |= x >> 2; x |= x >> 4; x |= x >> 8; x |= x >> 16; return x >> 1; } std::vector<uint> indices; std::vector<uint> inlabel; std::vector<uint> ascendant; std::vector<int> head; }; int main() { int N; while(~scanf("%d", &N)) { vector<vector<int> > g(N); for(int i = 0; i < N - 1; ++ i) { int u, v; scanf("%d%d", &u, &v), -- u, -- v; g[u].push_back(v); g[v].push_back(u); } tree_getorder(g, 0); SchieberVishkinLCA lca; lca.build(t_ord, t_parent, 0); vector<int> cnt(N); int Q; scanf("%d", &Q); rep(ii, Q) { int A; int B; scanf("%d%d", &A, &B), -- A, -- B; int C = lca.query(A, B); ++ cnt[A]; ++ cnt[B]; if(C != 0) -- cnt[t_parent[C]]; -- cnt[C]; } for(int ix = (int)t_ord.size() - 1; ix > 0; -- ix) { int i = t_ord[ix], p = t_parent[i]; cnt[p] += cnt[i]; } ll ans = 0; rep(i, N) { int n = cnt[i]; ans += (ll)n * (n + 1) / 2; } printf("%lld\n", ans); } return 0; }