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Shiroha白羽的博客

Golang 踩坑 —— interface 为参数的时候传 nil 指针 Codeforces Round 925 (Div. 3) Codeforces Round 924 (Div. 2) Codeforces Round 923 (Div. 3) Codeforces Round 922 (Div. 2) Codeforces Round 921 (Div. 2) Educational Codeforces Round 161 (Rated for Div. 2) Codeforces Round 920 (Div. 3) Codeforces Round 919 (Div. 2) Hello 2024 Good Bye 2023 Codeforces Round 918 (Div. 4) 个人备份的常用 macOS 清理命令 Codeforces Round 917 (Div. 2) Pinely Round 3 (Div. 1 + Div. 2) Educational Codeforces Round 160 (Rated for Div. 2) Codeforces Round 915 (Div. 2) Codeforces Round 914 (Div. 2) Codeforces Round 913 (Div. 3) Educational Codeforces Round 159 (Rated for Div. 2) Codeforces Round 912 (Div. 2) Codeforces Round 911 (Div. 2) CodeTON Round 7 (Div. 1 + Div. 2, Rated, Prizes!) Educational Codeforces Round 158 (Rated for Div. 2) Codeforces Round 910 (Div. 2) Codeforces Round 908 (Div. 2) Educational Codeforces Round 157 (Rated for Div. 2) C++自定义的字面量 Codeforces Round 907 (Div. 2) Codeforces Round 916 (Div. 3)
Codeforces Round 909 (Div. 3)
Shiroha · 2024-01-07 · via Shiroha白羽的博客

$$\begin{cases} & (2^{b_i})^{2^{b_j}} & = & (2^{b_j})^{2^{b_i}} \\ \Rightarrow & 2^{b_i \times 2^{b_j}} & = & 2^{b_j \times 2^{b_i}} \\ \Rightarrow & b_i \times 2^{b_j} & = & b_j \times 2^{b_i} \\ \Rightarrow & \frac{b_i}{b_j} & = & \frac{2^{b_i}}{2^{b_j}} \\ \Rightarrow & \frac{b_i}{b_j} & = & 2^{b_i - b_j} \end{cases}$$

$$\begin{cases} & \frac{b_j + x}{b_j} & = & 2^x \\ \Rightarrow & b_j + x & = & b_j \times 2^x \\ \Rightarrow & x & = & b_j \times (2^x - 1) \\ \Rightarrow & b_j & = & \frac{x}{2^x - 1} \\ \end{cases}$$

如果说此时在遍历到某个节点 $n$,这个节点在上面的数组 $p$ 的位置是 $m$,且这个 $m$ 恰好出现在了它的父节点的某个询问中,即父节点询问的区间包含 $m$
那么这个父节点的这个询问就是成功的,命中的。

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void solve() {
int _;
cin >> _;
for (int ts = 0; ts < _; ++ts) {
int n, q;
cin >> n >> q;
struct node {
int v, n;
};
vector<node> edge(n * 2 - 2);
vector<int> head(n + 1, -1), pos(n + 1);;
vector<vector<tuple<int, int, int>>> query(n + 1);
vector<pair<int, int>> qs;
for (int i = 0; i < n - 1; ++i) {
int u, v;
cin >> u >> v;
edge[i << 1] = {v, head[u]};
edge[i << 1 | 1] = {u, head[v]};
head[u] = i << 1;
head[v] = i << 1 | 1;
}
for (int i = 1; i <= n; ++i) {
int tmp;
cin >> tmp;
pos[tmp] = i;
}
for (int i = 0; i < q; ++i) {
int l, r, x;
cin >> l >> r >> x;
qs.emplace_back(l, r);
query[x].emplace_back(l, r, i);
}

vector<set<int>> tree(n * 2 + 10);
auto get = [](const int l, const int r) {
return (l + r) | (l != r);
};
function<void(int, int, int, int, int)> _add = [&](const int l, const int r, const int x, const int y, const int v) {
const int mid = l + r >> 1;
if (x == l && y == r) {
tree[get(l, r)].insert(v);
return;
}

if (y <= mid) _add(l, mid, x, y, v);
else if (x > mid) _add(mid + 1, r, x, y, v);
else {
_add(l, mid, x, mid, v);
_add(mid + 1, r, mid + 1, y, v);
}
};
function<bool(int, int, int, int, int)> _del = [&](const int l, const int r, const int x, const int y, const int v) {
const int mid = l + r >> 1;
if (x == l && y == r) {
return tree[get(l, r)].erase(v) ? true : false;
}

if (y <= mid) return _del(l, mid, x, y, v);
if (x > mid) return _del(mid + 1, r, x, y, v);
return _del(l, mid, x, mid, v) && _del(mid + 1, r, mid + 1, y, v);
};
function<void(int, int, int, vector<int>&)> _find = [&](const int l, const int r, const int v, vector<int>& res) {
for (const auto& i: tree[get(l, r)]) res.push_back(i);
if (l == r) return;

if (const int mid = l + r >> 1; v <= mid) {
_find(l, mid, v, res);
} else if (v > mid) {
_find(mid + 1, r, v, res);
}
};

vector ans(q, false);
vector<int> res;
res.reserve(n);

function<void(int, int)> dfs = [&](const int u, const int p) {
for (const auto [l, r, i]: query[u]) _add(1, n, l, r, i);
_find(1, n, pos[u], res);
for (const auto& i: res) {
ans[i] = true;
_del(1, n, qs[i].first, qs[i].second, i);
}
res.clear();

for (int i = head[u]; ~i; i = edge[i].n) {
if (edge[i].v == p) continue;
dfs(edge[i].v, u);
}

for (const auto [l, r, i]: query[u]) if (_del(1, n, l, r, i)) ans[i] = false;
};

dfs(1, 0);

for (const auto& i: ans) cout << (i ? "YES" : "NO") << endl;
cout << endl;
}
}