結果
問題 | No.399 動的な領主 |
ユーザー |
|
提出日時 | 2018-02-22 02:32:09 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 82 ms / 2,000 ms |
コード長 | 3,608 bytes |
コンパイル時間 | 2,208 ms |
コンパイル使用メモリ | 205,060 KB |
最終ジャッジ日時 | 2025-01-05 08:25:31 |
ジャッジサーバーID (参考情報) |
judge4 / judge4 |
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ファイルパターン | 結果 |
---|---|
other | AC * 19 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:106:30: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 106 | scanf("%d%d", &u, &v), -- u, -- v; | ~~~~~^~~~~~~~~~~~~~~~ main.cpp:115:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 115 | scanf("%d", &Q); | ~~~~~^~~~~~~~~~ main.cpp:118:30: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 118 | scanf("%d%d", &A, &B), -- A, -- B; | ~~~~~^~~~~~~~~~~~~~~~
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);elset_parent[i] = c;}}}using uint32 = unsigned int;class SchieberVishkinLCA {public:void build(const std::vector<int> &preorder, const std::vector<int> &parents, int root) {const int n = preorder.size();indices.resize(n);ascendant.resize(n);head.resize(n);for (int i = 0; i < n; ++i) indices[preorder[i]] = i + 1;inlabel.assign(indices.begin(), indices.end());for (int i = n - 1; i > 0; --i) {int v = preorder[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 = preorder[i], p = parents[v];ascendant[v] = ascendant[p] | lowbit(inlabel[v]);}head[0] = root;for (int i = 1; i < n; ++i) {int v = preorder[i], p = parents[v];if (inlabel[v] != inlabel[p]) head[indices[v] - 1] = p;else head[indices[v] - 1] = head[indices[p] - 1];}}int query(int u, int v) const {uint32 Iv = inlabel[v], Iu = inlabel[u];uint32 hIv = lowbit(Iv), hIu = lowbit(Iu);uint32 mask = highbit((Iv ^ Iu) | hIv | hIu) - 1;uint32 j = lowbit(ascendant[v] & ascendant[u] & ~mask);int x, y;if (j == hIv) x = v;else {mask = highbit(ascendant[v] & (j - 1)) - 1;x = head[(indices[v] & ~mask | (mask + 1)) - 1];}if (j == hIu) y = u;else {mask = highbit(ascendant[u] & (j - 1)) - 1;y = head[(indices[u] & ~mask | (mask + 1)) - 1];}return indices[x] < indices[y] ? x : y;}private:static uint32 lowbit(uint32 x) {return x & ~x + 1; // x & (-x) or x & (x ^ (x - 1))}static uint32 highbit(uint32 x) {return 1u << (31 - __builtin_clz(x));}std::vector<uint32> indices;std::vector<uint32> inlabel;std::vector<uint32> 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;}