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Segment Tree Merging & Splitting

Segment tree merging and splitting are common techniques for segment trees, often used in scenarios where weight segment trees maintain multisets.

For example, if certain nodes on a tree have several operations, and it is necessary to pass information from child nodes to parent nodes from bottom to top, while the information at a single node can be conveniently maintained using a segment tree, the technique of segment tree merging can be applied to control the overall complexity.

Segment Tree Merging

Process

As the name suggests, segment tree merging refers to constructing a new segment tree where each node is the result of merging the corresponding nodes of two original segment trees. It is often used to maintain information on trees or graphs.

Obviously, we cannot actually build a full new segment tree every time, so we need to use the dynamic segment tree mentioned above.

The process of segment tree merging is essentially quite brute-force:

Assume the two segment trees are A and B, and we start merging recursively from node 1.

When recursing to a certain node, if the corresponding node in tree A or tree B is empty, directly return the corresponding node from the other tree. This utilizes the characteristic of dynamic segment trees.

If recursing to a leaf node, we merge the corresponding nodes from the two trees.

Finally, update the current node based on its child nodes and return.

Complexity of Segment Tree Merging

Obviously, for two full segment trees, the complexity of a single merge operation is \(O(n)\). However, in practice, weight segment trees are often used, and the total number of nodes in all segment trees that need to be merged is not much larger than the scale of \(n\). Moreover, merging generally does not repeatedly merge a certain segment tree, so the total number of nodes we ultimately add is roughly \(n\log n\) level. Thus, the total complexity of merging all segment trees is \(O(n\log n)\) level. Of course, in some cases, mergeable heaps may be a better choice.

Implementation

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int merge(int a, int b, int l, int r) {
  if (!a) return b;
  if (!b) return a;
  if (l == r) {
    // do something...
    return a;
  }
  int mid = (l + r) >> 1;
  tr[a].l = merge(tr[a].l, tr[b].l, l, mid);
  tr[a].r = merge(tr[a].r, tr[b].r, mid + 1, r);
  pushup(a);
  return a;
}

Example

luogu P4556 [Vani Has a Date] Rainy Tail / 【Template】Segment Tree Merging
Solution Idea

A template problem for segment tree merging. Use difference to convert tree modifications into single-point modifications, then perform a bottom-up dfs to merge segment trees and calculate the answer.

Reference Code 
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#include <iostream>
#include <vector>
using namespace std;
int n, fa[100005][22], dep[100005], rt[100005];
int sum[5000005], cnt = 0, res[5000005], ls[5000005], rs[5000005];
int m, ans[100005];
vector<int> v[100005];

void update(int x) {
  if (sum[ls[x]] < sum[rs[x]]) {
    res[x] = res[rs[x]];
    sum[x] = sum[rs[x]];
  } else {
    res[x] = res[ls[x]];
    sum[x] = sum[ls[x]];
  }
}

int merge(int a, int b, int x, int y) {
  if (!a) return b;
  if (!b) return a;
  if (x == y) {
    sum[a] += sum[b];
    return a;
  }
  int mid = (x + y) >> 1;
  ls[a] = merge(ls[a], ls[b], x, mid);
  rs[a] = merge(rs[a], rs[b], mid + 1, y);
  update(a);
  return a;
}

int add(int id, int x, int y, int co, int val) {
  if (!id) id = ++cnt;
  if (x == y) {
    sum[id] += val;
    res[id] = co;
    return id;
  }
  int mid = (x + y) >> 1;
  if (co <= mid)
    ls[id] = add(ls[id], x, mid, co, val);
  else
    rs[id] = add(rs[id], mid + 1, y, co, val);
  update(id);
  return id;
}

void initlca(int x) {
  for (int i = 0; i <= 20; i++) fa[x][i + 1] = fa[fa[x][i]][i];
  for (int i : v[x]) {
    if (i == fa[x][0]) continue;
    dep[i] = dep[x] + 1;
    fa[i][0] = x;
    initlca(i);
  }
}

int lca(int x, int y) {
  if (dep[x] < dep[y]) swap(x, y);
  for (int d = dep[x] - dep[y], i = 0; d; d >>= 1, i++)
    if (d & 1) x = fa[x][i];
  if (x == y) return x;
  for (int i = 20; i >= 0; i--)
    if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
  return fa[x][0];
}

void cacl(int x) {
  for (int i : v[x]) {
    if (i == fa[x][0]) continue;
    cacl(i);
    rt[x] = merge(rt[x], rt[i], 1, 100000);
  }
  ans[x] = res[rt[x]];
  if (sum[rt[x]] == 0) ans[x] = 0;
}

int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  initlca(1);
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    rt[a] = add(rt[a], 1, 100000, c, 1);
    rt[b] = add(rt[b], 1, 100000, c, 1);
    int t = lca(a, b);
    rt[t] = add(rt[t], 1, 100000, c, -1);
    rt[fa[t][0]] = add(rt[fa[t][0]], 1, 100000, c, -1);
  }
  cacl(1);
  for (int i = 1; i <= n; i++) cout << ans[i] << endl;
  return 0;
}

Segment Tree Splitting

Process

Segment tree splitting is essentially the inverse process of segment tree merging. Segment tree splitting is only applicable to ordered sequences; it is meaningless for unordered sequences and is commonly used in dynamic weight segment trees.

Note that when both splitting and merging exist, we must recycle nodes during merging to avoid potential node duplication issues during splitting.

Split the interval \([l,r]\) from a segment tree covering \([1,N]\) to form a new tree:

Start recursive splitting from node 1. If the node does not exist or the represented interval \([s,t]\) does not intersect with \([l,r]\), backtrack directly.

When \([s,t]\) intersects with \([l,r]\), a new node needs to be created.

When \([s,t]\) is completely contained within \([l,r]\), the current node needs to be directly attached under the new tree, and the old edge is disconnected.

Complexity of Segment Tree Splitting

It can be observed that at most \(\log n\) edges will be disconnected, so the time complexity of each split operation is \(O(\log n)\), equivalent to the complexity of a range query.

Implementation

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void split(int &p, int &q, int s, int t, int l, int r) {
  if (t < l || r < s) return;
  if (!p) return;
  if (l <= s && t <= r) {
    q = p;
    p = 0;
    return;
  }
  if (!q) q = New();
  int m = s + t >> 1;
  if (l <= m) split(ls[p], ls[q], s, m, l, r);
  if (m < r) split(rs[p], rs[q], m + 1, t, l, r);
  push_up(p);
  push_up(q);
}

Example

P5494 【Template】Segment Tree Splitting
Solution Idea

A template problem for segment tree splitting. Split the interval \([x,y]\).

  • Merge tree \(t\) into tree \(p\): a single merge operation.

  • Insert \(x\) copies of \(q\) into tree \(p\): single-point modification.

  • Query the number of elements in \([x,y]\): range sum.

  • Query the \(k\)-th smallest element.

Reference Code 
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#include <iostream>
using namespace std;
constexpr int N = 2e5 + 10;
int n, m;
int idx = 1;
long long sum[N << 5];
int ls[N << 5], rs[N << 5], root[N << 2], rub[N << 5], cnt, tot;

// Memory allocation and recycling
int New() { return cnt ? rub[cnt--] : ++tot; }

void Del(int &p) {
  ls[p] = rs[p] = sum[p] = 0;
  rub[++cnt] = p;
  p = 0;
}

void push_up(int p) { sum[p] = sum[ls[p]] + sum[rs[p]]; }

void build(int s, int t, int &p) {
  if (!p) p = New();
  if (s == t) {
    cin >> sum[p];
    return;
  }
  int m = (s + t) >> 1;
  build(s, m, ls[p]);
  build(m + 1, t, rs[p]);
  push_up(p);
}

// Single point update
void update(int x, int c, int s, int t, int &p) {
  if (!p) p = New();
  if (s == t) {
    sum[p] += c;
    return;
  }
  int m = (s + t) >> 1;
  if (x <= m)
    update(x, c, s, m, ls[p]);
  else
    update(x, c, m + 1, t, rs[p]);
  push_up(p);
}

// Merge
int merge(int p, int q, int s, int t) {
  if (!p || !q) return p + q;
  if (s == t) {
    sum[p] += sum[q];
    Del(q);
    return p;
  }
  int m = (s + t) >> 1;
  ls[p] = merge(ls[p], ls[q], s, m);
  rs[p] = merge(rs[p], rs[q], m + 1, t);
  push_up(p);
  Del(q);
  return p;
}

// Split
void split(int &p, int &q, int s, int t, int l, int r) {
  if (t < l || r < s) return;
  if (!p) return;
  if (l <= s && t <= r) {
    q = p;
    p = 0;
    return;
  }
  if (!q) q = New();
  int m = (s + t) >> 1;
  if (l <= m) split(ls[p], ls[q], s, m, l, r);
  if (m < r) split(rs[p], rs[q], m + 1, t, l, r);
  push_up(p);
  push_up(q);
}

long long query(int l, int r, int s, int t, int p) {
  if (!p) return 0;
  if (l <= s && t <= r) return sum[p];
  int m = (s + t) >> 1;
  long long ans = 0;
  if (l <= m) ans += query(l, r, s, m, ls[p]);
  if (m < r) ans += query(l, r, m + 1, t, rs[p]);
  return ans;
}

int kth(int s, int t, int k, int p) {
  if (s == t) return s;
  int m = (s + t) >> 1;
  long long left = sum[ls[p]];
  if (k <= left)
    return kth(s, m, k, ls[p]);
  else
    return kth(m + 1, t, k - left, rs[p]);
}

int main() {
  cin >> n >> m;
  build(1, n, root[1]);
  while (m--) {
    int op;
    cin >> op;
    if (!op) {
      int p, x, y;
      cin >> p >> x >> y;
      split(root[p], root[++idx], 1, n, x, y);
    } else if (op == 1) {
      int p, t;
      cin >> p >> t;
      root[p] = merge(root[p], root[t], 1, n);
    } else if (op == 2) {
      int p, x, q;
      cin >> p >> x >> q;
      update(q, x, 1, n, root[p]);
    } else if (op == 3) {
      int p, x, y;
      cin >> p >> x >> y;
      cout << query(x, y, 1, n, root[p]) << endl;
    } else {
      int p, k;
      cin >> p >> k;
      if (sum[root[p]] < k)
        cout << -1 << endl;
      else
        cout << kth(1, n, k, root[p]) << endl;
    }
  }
}

Exercises

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