結果
| 問題 |
No.399 動的な領主
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2018-02-22 02:29:46 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 3,609 bytes |
| コンパイル時間 | 2,204 ms |
| コンパイル使用メモリ | 205,036 KB |
| 最終ジャッジ日時 | 2025-01-05 08:25:18 |
|
ジャッジサーバーID (参考情報) |
judge4 / judge3 |
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| ファイルパターン | 結果 |
|---|---|
| other | RE * 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);
else
t_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();
inlabel.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;
}