点分治

Divide and conquer with node

点分治, 就是选取适当的根节点, 把树上关于路径的问题分成两类考虑, 即经过根的路径和不经过根的路径, 在求解时只考虑经过根的路径, 然后对于根的每个儿子的子树递归求解.

复杂度和层数还有每层的点数有关, 因为每层有 \(O(n)\) 个点, 所以我们要尽量减少层数. 如果选择树的重心, 那么最坏情况下层数是 \(O(\log n)\). 因此对于每个子问题都选择重心作为根.

模板题

给一棵边带权的树, 求是否存在距离为 \(k\) 的点对.

如果不会点分治, 可以使用 DSU on Tree 在 \(O(n\log n)\) 的时间内解决问题, 而且常数更小. 不过我们为了学习点分治, 只能先放下这个方法.

对于只经过根的点对是否存在, 只需要 DFS 一遍, 把所有点的深度存到 set 里即可, 实现 \(O(n\log^2 n)\) 的算法, 也可以使用哈希表, 达到 \(O(n\log n)\) 的水平.

由于是 \(m\) 次询问, 我们可以把询问都存下来, 统一回答, 可以减少常数.

const unsigned long long Mod(1000003), Base(41);
unsigned Q[105], m, n;
unsigned A, B, C, D, t;
unsigned Cnt(0), Tmp(0), Mn(0), Total;
unsigned Hash[1010005], Stack[10005], STop(0);
unsigned Buffer[10005], BTop(0);
bitset<105> Ans;
struct Node {
  vector <pair<Node*, unsigned> > To;
  unsigned Size, MxSize, Dep;
  char NAva;
}N[10005], * Heavy;
inline unsigned Find(unsigned x) {
  unsigned Pos(x * Base % Mod);
  while (Hash[Pos] && (Hash[Pos] ^ x)) ++Pos;
  return Pos;
}
inline void DFS1(Node* x, Node* Fa) {
  x->Size = 1, x->MxSize = 0;
  for (auto i : x->To) if ((i.first != Fa) && (!(i.first->NAva)))
    DFS1(i.first, x), x->Size += i.first->Size, x->MxSize = max(x->MxSize, i.first->Size);
}
inline void DFS2(Node* x, Node* Fa) {
  unsigned MnMx(max(x->MxSize, Total - x->Size));
  if (MnMx < Mn) Mn = MnMx, Heavy = x;
  for (auto i : x->To) if ((i.first != Fa) && (!(i.first->NAva))) DFS2(i.first, x);
}
inline void DFS3(Node* x, Node* Fa) {
  unsigned Pos;
  for (auto i : x->To) if ((i.first != Fa) && (!(i.first->NAva))) {
    Buffer[++BTop] = (i.first->Dep = x->Dep + i.second);
    for (unsigned j(1); j <= m; ++j) if ((!Ans[j]) && (Q[j] >= i.first->Dep)) {
      if (i.first->Dep == Q[j]) { Ans[j] = 1; continue; }
      Pos = Find(Q[j] - i.first->Dep); if (Hash[Pos] == (Q[j] - i.first->Dep)) Ans[j] = 1;
    }
    DFS3(i.first, x);
  }
}
inline void Solve(Node* x) {
  DFS1(x, NULL), Heavy = NULL, Mn = 0x3f3f3f3f, Total = x->Size, DFS2(x, NULL);
  unsigned Pos;
  Heavy->Dep = 0;
  for (auto i : Heavy->To) if (!(i.first->NAva)) {
    Buffer[++BTop] = (i.first->Dep = Heavy->Dep + i.second);
    for (unsigned j(1); j <= m; ++j) if ((!Ans[j]) && (Q[j] >= i.first->Dep)) {
      if (i.first->Dep == Q[j]) { Ans[j] = 1; continue; }
      Pos = Find(Q[j] - i.first->Dep); if (Hash[Pos] == (Q[j] - i.first->Dep)) Ans[j] = 1;
    }
    DFS3(i.first, Heavy);
    while (BTop) { Pos = Find(Buffer[BTop]); if (!Hash[Pos]) Hash[Pos] = Buffer[BTop], Stack[++STop] = Pos; --BTop; }
  }
  Heavy->NAva = 1;
  while (STop) Hash[Stack[STop--]] = 0;
  for (auto i : Heavy->To) if (!(i.first->NAva)) Solve(i.first);
}
signed main() {
  n = RD(), m = RD();
  for (unsigned i(1); i < n; ++i) {
    A = RD(), B = RD(), C = RD();
    N[A].To.push_back(make_pair(N + B, C));
    N[B].To.push_back(make_pair(N + A, C));
  }
  for (unsigned i(1); i <= m; ++i) Q[i] = RD();
  Solve(N + 1);
  for (unsigned i(1); i <= m; ++i)printf(Ans[i] ? "AYE\n" : "NAY\n");
  return Wild_Donkey;
}