結果
| 問題 |
No.399 動的な領主
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2019-10-27 16:33:05 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 74 ms / 2,000 ms |
| コード長 | 3,922 bytes |
| コンパイル時間 | 2,108 ms |
| コンパイル使用メモリ | 204,684 KB |
| 最終ジャッジ日時 | 2025-01-08 02:00:52 |
|
ジャッジサーバーID (参考情報) |
judge2 / judge5 |
(要ログイン)
| 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:109:12: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
109 | scanf("%d%d", &u, &v), -- u, -- v;
| ~~~~~^~~~~~~~~~~~~~~~
main.cpp:118:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
118 | scanf("%d", &Q);
| ~~~~~^~~~~~~~~~
main.cpp:121:12: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
121 | 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);
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;
}