树上启发式合并 学习记录

算法讲解163【挺难】树上启发式合并的原理和相关题目_哔哩哔哩_bilibili

适用于统计一个树中多个子树的信息的题目

适用于没有修改操作的题目

原理: 

对于每个轻儿子的子树先遍历统计他的答案, 再消除这个子树的影响, 然后遍历重儿子, 统计答案, 再遍历一遍轻儿子统计答案

因为对于每个点, 他会被再遍历一次, 那么他所处的集合就会倍增一次, 所以他被遍历的次数不超过log(n),  总体复杂度O(nlogn)

题目

U41492 树上数颜色 - 洛谷

#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
// 不能使用0下标
// 传n + 1 使用下标 1~n
struct HLD {HLD(const int &n) : n(n), edge(n) {fa = dep = son = sz = top = dfn = out = rnk = std::vector<int>(n);}void addEdge(const int &x, const int &y) {edge[x].push_back(y);edge[y].push_back(x);}void dfs1(int u, int p) {sz[u] = 1;fa[u] = p;dep[u] = dep[p] + 1;for(const auto &v : edge[u]) {if(v == fa[u]) continue;dfs1(v, u);sz[u] += sz[v];if(sz[son[u]] < sz[v]) {son[u] = v;}}}void dfs2(int u, int t, int &dfncnt) {dfncnt++;dfn[u] = dfncnt;rnk[dfncnt] = u;top[u] = t;if(son[u] == 0) return;dfs2(son[u], t, dfncnt);for(const auto &v : edge[u]) {if(v == fa[u] || v == son[u]) continue;dfs2(v, v, dfncnt);}}void work(int root = 1) {int dfncnt = 0;dfs1(root, 0);dfs2(root, root, dfncnt);}int lca(int u, int v) {while(top[u] != top[v]) {if(dep[top[u]] < dep[top[v]]) {std::swap(u, v);}u = fa[top[u]];}return (dep[u] < dep[v] ? u : v);}int dis(int x, int y) {return dep[x] + dep[y] - 2 * dep[lca(x, y)];}// int kth(int id, int k) {//  if(k > dep[id]) return 0;//  while(dep[id] - dep[top[id]] + 1 <= k) {//      k -= (dep[id] - dep[top[id]] + 1);//      id = fa[top[id]];//  }//  return rnk[dfn[id] - k];// }vector<vector<int>> edge;vector<int> fa, dep, son, sz, top, dfn, out, rnk;int n;
};void solve() {int n;cin >> n;HLD tree(n + 1);vector<vector<int>> edge(n + 1);for (int i = 1; i < n; i++) {int u, v;cin >> u >> v;edge[u].push_back(v);edge[v].push_back(u);tree.addEdge(u, v);}vector<int> a(n + 1);for (int i = 1; i <= n; i++) cin >> a[i];vector<int> cnt(n + 1);int dif = 0;vector<int> ans(n + 1);tree.work(1);auto &dfn = tree.dfn;auto &rnk = tree.rnk;auto &son = tree.son;auto &sz = tree.sz;auto dfs = [&] (auto self, int u, int p, int keep) -> void {for (auto v : edge[u]) {if (v == p || v == son[u]) continue;self(self, v, u, 0);}if (son[u]) self(self, son[u], u, 1);cnt[a[u]]++;if (cnt[a[u]] == 1) dif++;for (auto v : edge[u]) {if (v == p || v == son[u]) continue;for (int i = dfn[v]; i <= dfn[v] + sz[v] - 1; i++) {cnt[a[rnk[i]]]++;if (cnt[a[rnk[i]]] == 1) dif++;}}ans[u] = dif;if (!keep) {dif = 0;for (int i = dfn[u]; i <= dfn[u] + sz[u] - 1; i++) {cnt[a[rnk[i]]]--;}}};dfs(dfs, 1, 0, 1);int m;cin >> m;while(m--) {int x;cin >> x;cout << ans[x] << '\n';}}int main() {std::ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = 1;//cin >> T;while (T--) solve();return 0;
}