結果

問題 No.2337 Equidistant
ユーザー Kude
提出日時 2023-06-02 22:21:16
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 218 ms / 4,000 ms
コード長 4,322 bytes
コンパイル時間 2,381 ms
コンパイル使用メモリ 225,560 KB
最終ジャッジ日時 2025-02-13 18:50:26
ジャッジサーバーID
(参考情報)
judge3 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 1
other AC * 28
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
namespace {
#pragma GCC diagnostic ignored "-Wunused-function"
#include<atcoder/all>
#pragma GCC diagnostic warning "-Wunused-function"
using namespace std;
using namespace atcoder;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
struct HLD {
const vector<vector<int>>& to;
int root, n;
vector<int> sz, parent, depth, idx, ridx, head, inv;
HLD(const vector<vector<int>>& to, int root=0)
: to(to), root(root), n(to.size()),
sz(n), parent(n), depth(n), idx(n), ridx(n), head(n), inv(n) {
init_tree_data(root, -1, 0);
int nxt = 0;
assign_idx(root, root, nxt);
}
void init_tree_data(int u, int p, int d) {
parent[u] = p;
depth[u] = d;
int s = 1;
for (int v: to[u]) if (v != p) {
init_tree_data(v, u, d+1);
s += sz[v];
}
sz[u] = s;
}
void assign_idx(int u, int h, int& nxt, int p=-1) {
head[u] = h;
idx[u] = nxt;
inv[nxt] = u;
nxt++;
int heaviest = -1;
int mxweight = 0;
for (int v: to[u]) if (v != p) {
if (sz[v] > mxweight) {
heaviest = v;
mxweight = sz[v];
}
}
if (heaviest != -1) {
assign_idx(heaviest, h, nxt, u);
for (int v: to[u]) if (v != p && v != heaviest) {
assign_idx(v, v, nxt, u);
}
}
ridx[u] = nxt;
}
int lca(int u, int v) {
while (head[u] != head[v]) {
if (depth[head[u]] > depth[head[v]]) {
u = parent[head[u]];
} else {
v = parent[head[v]];
}
}
return depth[u] < depth[v] ? u : v;
}
// returns reference to tuple of (path fragments from x upto lca (excluding lca), those from y, lca)
// storage of retval is reused to avoid creating short vectors on each query
tuple<vector<pair<int, int>>, vector<pair<int, int>>, int> paths_res;
auto& paths(int x, int y) {
auto& [x_paths, y_paths, lca] = paths_res;
x_paths.clear();
y_paths.clear();
while (head[x] != head[y]) {
int hx = head[x], hy = head[y];
if (depth[hx] > depth[hy]) {
x_paths.emplace_back(x, hx); x = parent[hx];
} else {
y_paths.emplace_back(y, hy); y = parent[hy];
}
}
if (depth[x] > depth[y]) {
x_paths.emplace_back(x, inv[idx[y] + 1]); x = y;
} else if (depth[x] < depth[y]) {
y_paths.emplace_back(y, inv[idx[x] + 1]); y = x;
}
lca = x;
return paths_res;
}
int dist(int u, int v) {
int w = lca(u, v);
return depth[u] + depth[v] - 2 * depth[w];
}
template <class F> int max_ancestor(int v, F f) {
if (!f(v)) return -1;
int hv = head[v];
int p = parent[hv];
while (p != -1 && f(p)) {
v = p;
hv = head[v];
p = parent[hv];
}
int il = idx[hv] - 1, ir = idx[v];
while (ir - il > 1) {
int ic = (il + ir) / 2;
(f(inv[ic]) ? ir : il) = ic;
}
return inv[ir];
}
int ascend(int v, int k) {
assert(depth[v] >= k);
int td = depth[v] - k;
int hv = head[v];
while (depth[hv] > td) {
v = parent[hv];
hv = head[v];
}
int rest = depth[v] - td;
return inv[idx[v] - rest];
}
};
} int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, q;
cin >> n >> q;
VVI to(n);
rep(_ , n - 1) {
int u, v;
cin >> u >> v;
u--, v--;
to[u].emplace_back(v);
to[v].emplace_back(u);
}
HLD hld(to);
rep(_, q) {
int s, t;
cin >> s >> t;
s--, t--;
if (hld.depth[s] < hld.depth[t]) swap(s, t);
int d = hld.dist(s, t);
if (d % 2) {
cout << 0 << '\n';
continue;
}
int c = hld.ascend(s, d / 2);
int ans = n;
for(int v : {s, t}) {
int l = hld.lca(v, c);
if (l == c) {
int x = hld.ascend(v, d / 2 - 1);
ans -= hld.sz[x];
} else {
ans -= n - hld.sz[c];
}
}
cout << ans << '\n';
}
}
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