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Problem solution · C++

Dynamic Tree Vertex Add Subtree Sum

Dynamic Tree Vertex Add Subtree Sum: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to NyaanNyaan Competitive Programming Library.

Technique
Direct simulation
Source
NyaanNyaan Competitive Programming Library
Length
53 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Dynamic Tree Vertex Add Subtree Sum, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 53 lines of C++ from the credited upstream file yosupo-dynamic-tree-vertex-add-subtree-sum.test.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from NyaanNyaan Competitive Programming Library by NyaanNyaan and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeDynamic Tree Vertex Add Subtree Sum · C++C++
Use this to learn the idea, then write your own version.
#define PROBLEM "https://judge.yosupo.jp/problem/dynamic_tree_vertex_add_subtree_sum" #include "../../template/template.hpp"//#include "../../lct/link-cut-tree-subtree.hpp" using namespace Nyaan; using T = ll;T f(ll a, ll b) { return a + b; }T finv(ll a, ll b) { return a - b; } #include "../../misc/fastio.hpp"void Nyaan::solve() {  using LCT = LinkCutTreeSubtreeQuery<T, f, finv>;  LCT lct;   int N, Q;  rd(N, Q);   vector<LCT::Ptr> vs(N);  rep(i, N) {    int a;    rd(a);    vs[i] = lct.my_new(a);  }   for (int i = 1; i < N; i++) {    int a, b;    rd(a, b);    lct.link(vs[a], vs[b]);  }  while (Q--) {    int cmd;    rd(cmd);    if (cmd == 0) {      int u, v, w, x;      rd(u, v, w, x);      lct.cut(vs[u], vs[v]);      lct.link(vs[w], vs[x]);    } else if (cmd == 1) {      int u, x;      rd(u, x);      lct.set_key(vs[u], lct.get_key(vs[u]) + x);    } else {      int v, p;      rd(v, p);      lct.evert(vs[p]);      wtn(lct.subtree(vs[v]));    }  }}

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