老年 ACMer 尝试对抗阿尔茨海默病（
图论
DFS序 \(O(n \log n)\) - \(O(1)\) Lca
考虑点\(u\)，\(v\) 及其 \(Lca\) 点\(l\)，不妨设 $dfn_u \lt dfn_v $，那么有 \(dfs\) 序从 \(l\) 到 \(u\) 递增，此后回到 \(l\) 后再向着 \(v\)递增
所以，按\(dfs\)序构成的序列中，\(dfn_u\)到\(dfn_v\)这一段中深度最小的点一定是\(l\)在\(v\)方向直接连接的节点
所以可以直接使用\(ST\)表维护父节点\(dfs\)序的最小值即可
注意可能有\(u\)为\(v\)的祖先的情况，因此\(ST\)表询问区间该为\([dfn_u+1,dfn_v]\)进行规避
P3379 【模板】最近公共祖先（LCA）
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int mindfn(int x, int y) {if (dfn[x] < dfn[y]) return x;return y;
}struct ST {int st[MAXN][25];void init(int n) {int t = std::__lg(n) + 1;for (int i = 1; i <= n; i++) st[dfn[i]][0] = fa[i];for (int j = 1; j <= t; j++) {for (int i = 1; i + (1 << (j - 1)) <= n; i++) {st[i][j] = mindfn(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);}}}int query(int l, int r) {int t = std::__lg(r - l + 1);return mindfn(st[l][t], st[r - (1 << t) + 1][t]);}
}st;auto Lca = [&](int x, int y) {if (dfn[x] > dfn[y]) std::swap(x, y);return st.query(dfn[x] + 1, dfn[y]);
};
Dsu on tree
当涉及子树问题时，对于父亲相同的子树，如果在计算一个的答案时保留了其他子树的信息，统计会出现混乱。但是这种保留对于父亲子树的统计有利
于是考虑对于每个子树，先计算它轻儿子子树中的答案，并且不保留。最后计算重儿子的答案并保留，将轻儿子的答案往上合并
对于每个节点，每当其到根节点有一条轻链时会被暴力合并一次，由树链剖分的性质可知最多会被合并 \(O(\log n)\)次
CF600E Lomsat gelral
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#include <bits/stdc++.h>#define ll long longconst int MAXN = 1e5 + 5;int n, a[MAXN], fa[MAXN], siz[MAXN], son[MAXN];
std::vector <int> G[MAXN];
ll ans[MAXN];
int buc[MAXN], mx, S;
ll sum;int main() {std::ios::sync_with_stdio(0);std::cin.tie(0);std::cout.tie(0);int n;std::cin >> n;for (int i = 1; i <= n; i++) std::cin >> a[i];for (int i = 1; i < n; i++) {int u, v;std::cin >> u >> v;G[u].push_back(v);G[v].push_back(u);}auto dfs1 = [&](int u, int f, auto self) -> void {fa[u] = f; siz[u] = 1;for (const auto &v : G[u]) {if (v == f) continue;self(v, u, self);siz[u] += siz[v];if (siz[v] > siz[son[u]]) son[u] = v;}};dfs1(1, 0, dfs1);auto calc = [&](int u, int delta, auto self) -> void {buc[a[u]] += delta;if (buc[a[u]] > mx) {mx = buc[a[u]];sum = a[u];}else if (buc[a[u]] == mx) {sum += a[u];}for (const auto &v : G[u]) {if (v == fa[u] || v == S) continue;self(v, delta, self);}};auto dfs2 = [&](int u, bool keep, auto self) -> void {for (const auto &v : G[u]) {if (v == fa[u] || v == son[u]) continue;self(v, 0, self);}if (son[u]) {self(son[u], 1, self);S = son[u];}calc(u, 1, calc);ans[u] = sum;S = 0;if (!keep) {calc(u, -1, calc);mx = 0; sum = 0;}};dfs2(1, 1, dfs2);for (int i = 1; i <= n; i++) {std::cout << ans[i] << " "; }std::cout << '\n';
}
点分治
对于树上的路径问题，我们考虑固定一个根的情况，答案分为两种：跨过根的路径(包括以根为端点)和根的子树内部的路径
显然后者有子问题结构，考虑分治。当每次的根都取子树重心时，有最少递归层数 \(O(\log n)\)
由于较小的递归层数，我们可以每一层都采取相对暴力的做法来统计跨过根的路径
注意每次向下分治时重新计算子树大小，不然会导致重心求解出错，时间复杂度假掉
P3806 【模板】点分治
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#include <bits/stdc++.h>const int MAXN = 1e4 + 5;std::vector <std::pair <int, int> > G[MAXN];
int qry[105];
bool ans[105];
std::bitset <10000005> exist;int main() {int n, q;std::cin >> n >> q;for (int i = 1; i < n; i++) {int u, v, w;std::cin >> u >> v >> w;G[u].push_back({v, w});G[v].push_back({u, w});}for (int i = 1; i <= q; i++) {std::cin >> qry[i];}std::vector <bool> vis(n + 1);std::vector <int> mx(n + 1), siz(n + 1);int rt;int node_cnt;auto Find_size = [&](int u, int f, auto self) -> void {siz[u] = 1;for (const auto &p : G[u]) {int v = p.first, w = p.second;if (v == f || vis[v]) continue;self(v, u, self);siz[u] += siz[v];}};auto Find_cent = [&](int u, int f, auto self) -> void {mx[u] = 0;for (const auto &p : G[u]) {int v = p.first, w = p.second;if (vis[v] || v == f) continue;self(v, u, self);mx[u] = std::max(mx[u], siz[v]);}mx[u] = std::max(mx[u], node_cnt - siz[u]);if (mx[u] < mx[rt]) rt = u;};std::vector <int> dis;auto Find_dis = [&](int u, int f, int d, auto self) -> void {if (d > 10000000) return;dis.push_back(d);for (const auto & p : G[u]) {int v = p.first, w = p.second;if (v == f || vis[v]) continue;self(v, u, d + w, self);}};std::vector <int> used;auto calc = [&](int u, int d, auto self) -> void {Find_dis(u, u, d, Find_dis);for (const auto &x : dis) {for (int i = 1; i <= q; i++) {if (qry[i] - x >= 0) ans[i] |= exist[qry[i] - x];}}for (const auto &x : dis) {exist[x] = 1;used.push_back(x);}dis.clear();};auto solve = [&](int u, auto self) -> void {vis[u] = 1; exist[0] = 1;for (const auto &p : G[u]) {int v = p.first, w = p.second;if (vis[v]) continue;calc(v, w, calc);} for (const auto &x : used) exist[x] = 0;used.clear();for (const auto &p : G[u]) {int v = p.first;if (vis[v]) continue;mx[rt = 0] = 0x3f3f3f3f;Find_size(v, u, Find_size);node_cnt = siz[v];Find_cent(v, u, Find_cent);self(rt, self);}};mx[rt = 0] = 0x3f3f3f3f;node_cnt = n;Find_size(1, 0, Find_size);Find_cent(1, 0, Find_cent);solve(rt, solve);for (int i = 1; i <= q; i++) {if (ans[i]) {std::cout << "AYE" << '\n';}else std::cout << "NAY" << '\n';}
}
点分树
考虑利用点分治只会递归\(\log n\)层的优秀性质，每次递归的时候，子树的重心向当前根节点连边，就形成了一颗\(\log n\)的树
这棵树满足对于\(l = Lca(u, v)\)，必有 \(l\) 在原树 \(u\)到\(v\)的路径上
因此可以从在点分树上从\(u\)开始往上跳，假设当前跳到了\(l\)，那么对\(l\)除去\(u\)方向的子树的所有节点，都有一条在原树中经过\(l\)到达\(u\)的路径，可以采用数据结构进行统计
树高的性质保证了最多向上跳 \(\log n\) 次，时间复杂度正确
P6329 【模板】点分树 / 震波
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#include <bits/stdc++.h>const int MAXN = 1e5 + 5;int n, m, a[MAXN];
std::vector <int> G[MAXN];struct Sgt {struct Node {int ls, rs, sum;}t[MAXN * 80];int cnt;void update(int &p, int l, int r, int pos, int val) {if (!p) p = ++cnt;t[p].sum += val;if (l == r) return;int mid = (l + r) >> 1;if (pos <= mid) update(t[p].ls, l, mid, pos, val);else update(t[p].rs, mid + 1, r, pos, val);}int query(int p, int l, int r, int ql, int qr) {if (!p || ql > qr) return 0;if (ql <= l && r <= qr) return t[p].sum;int mid = (l + r) >> 1, ret = 0;if (ql <= mid) ret += query(t[p].ls, l, mid, ql, qr);if (qr > mid) ret += query(t[p].rs, mid + 1, r, ql, qr);return ret;}
}t1, t2;
int rt1[MAXN], rt2[MAXN];int lst;
int dfn[MAXN], fa[MAXN], dcnt, dep[MAXN];void dfs(int u, int f) {fa[u] = f; dfn[u] = ++dcnt; dep[u] = dep[f] + 1;for (const auto &v : G[u]) {if (v == f) continue;dfs(v, u);}
}struct ST {int st[MAXN][25];int dfnmin(int x, int y) {if (dfn[x] < dfn[y]) return x;return y;}void init() {int t = std::__lg(n) + 1;for (int i = 1; i <= n; i++) st[dfn[i]][0] = fa[i];for (int j = 1; j <= t; j++) {for (int i = 1; i + (1 << (j - 1)) <= n; i++) {st[i][j] = dfnmin(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);}}}int query(int l, int r) {int t = std::__lg(r - l + 1);return dfnmin(st[l][t], st[r - (1 << t) + 1][t]);}
}st;int Lca(int x, int y) {if (x == y) return x;if (dfn[x] > dfn[y]) std::swap(x, y);return st.query(dfn[x] + 1, dfn[y]);
}int dis(int x, int y) {int l = Lca(x, y);return dep[x] + dep[y] - 2 * dep[l];
}int fa2[MAXN], mx[MAXN], siz[MAXN];int main() {std::ios::sync_with_stdio(0);std::cin.tie(0);std::cout.tie(0);std::cin >> n >> m;for (int i = 1; i <= n; i++) std::cin >> a[i];for (int i = 1; i < n; i++) {int u, v;std::cin >> u >> v;G[u].push_back(v);G[v].push_back(u);}dfs(1, 0);st.init();std::vector <bool> vis(n + 1);int node_cnt, rt;auto Find_cent = [&](int u, int f, auto self) -> void {mx[u] = 0;for (const auto &v : G[u]) {if (vis[v] || v == f) continue;self(v, u, self);mx[u] = std::max(mx[u], siz[v]);}mx[u] = std::max(mx[u], node_cnt - siz[u]);if (mx[rt] > mx[u]) rt = u;};auto Find_size = [&](int u, int f, auto self) -> void {siz[u] = 1;for (const auto &v : G[u]) {if (vis[v] || v == f) continue;self(v, u, self);siz[u] += siz[v];}};auto build = [&](int u, auto self) -> void {vis[u] = 1; Find_size(u, -1, Find_size);for (const auto &v : G[u]) {if (vis[v]) continue;mx[rt = 0] = 0x3f3f3f3f;node_cnt = siz[v];Find_cent(v, u, Find_cent);fa2[rt] = u;self(rt, self);}};mx[rt = 0] = 0x3f3f3f3f;node_cnt = n;Find_cent(1, 0, Find_cent);fa2[rt] = 0;build(rt, build);auto query = [&](int x, int k) {int f = fa2[x], cur = x, ret = t1.query(rt1[x], 0, n, 0, k);while (f) {int d = k - dis(x, f);ret += (t1.query(rt1[f], 0, n, 0, d) - t2.query(rt2[cur], 0, n, 0, d));cur = fa2[cur];f = fa2[f];}return ret;};auto update = [&](int x, int y, bool flag) {int cur = x;while (cur) {if (flag) t1.update(rt1[cur], 0, n, dis(x, cur), -a[x]);t1.update(rt1[cur], 0, n, dis(x, cur), y);int f = fa2[cur];if (f) {if (flag) t2.update(rt2[cur], 0, n, dis(x, f), -a[x]);t2.update(rt2[cur], 0, n, dis(x, f), y);}cur = f;}};for (int i = 1; i <= n; i++) {update(i, a[i], 0);}while (m--) {int op, x, y;std::cin >> op >> x >> y;x ^= lst; y ^= lst;if (op == 0) {std::cout << (lst = query(x, y)) << '\n';} if (op == 1) {update(x, y, 1);a[x] = y;}}return 0;
}
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