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Problem solution · C++

Shortest Path in a Weighted Tree

Shortest Path in a Weighted Tree: a C++ solution using segment tree or range structure. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Segment tree or range structure
Source
walkccc LeetCode Solutions
Length
123 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

Segment tree or range structure

For Shortest Path in a Weighted Tree, the implementation stores interval information in a range-query data structure so updates and queries avoid rescanning the full input.

  1. Choose the aggregate stored for each interval or prefix.
  2. Build or initialize the structure from the input.
  3. Apply updates and combine the affected nodes to answer each query.

Code notes

  • 123 lines of C++ from the credited upstream file 3515.cpp.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 4 loop blocks detected, together with recursive traversal.

Complexity

Count the build once, then multiply the logarithmic update or query path by the number of operations.

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

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeShortest Path in a Weighted Tree · C++C++
Use this to learn the idea, then write your own version.
class LazySegmentTree { public:  explicit LazySegmentTree(int n) : n(n), tree(4 * n), lazy(4 * n) {}   // Updates the range [l, r] by adding val.  void addRange(int l, int r, int val) {    addRange(0, 0, n - 1, l, r, val);  }   // Returns the value at index i.  int query(int i) {    return query(0, 0, n - 1, i);  }  private:  const int n;       // the size of the input array  vector<int> tree;  // the segment tree  vector<int> lazy;  // the lazy propagation array   void push(int treeIndex, int lo, int hi) {    if (lazy[treeIndex] == 0)      return;    tree[treeIndex] += lazy[treeIndex];    if (lo != hi) {      lazy[2 * treeIndex + 1] += lazy[treeIndex];      lazy[2 * treeIndex + 2] += lazy[treeIndex];    }    lazy[treeIndex] = 0;  }   void addRange(int treeIndex, int lo, int hi, int l, int r, int val) {    push(treeIndex, lo, hi);    if (r < lo || l > hi)  // [lo, hi] lies completely outside [l, r].      return;    if (l <= lo && hi <= r) {  // [lo, hi] lies completely inside [l, r].      lazy[treeIndex] += val;      push(treeIndex, lo, hi);      return;    }    const int mid = (lo + hi) / 2;    addRange(2 * treeIndex + 1, lo, mid, l, r, val);    addRange(2 * treeIndex + 2, mid + 1, hi, l, r, val);  }   int query(int treeIndex, int lo, int hi, int i) {    push(treeIndex, lo, hi);    if (lo == hi)      return tree[treeIndex];    const int mid = (lo + hi) / 2;    if (i <= mid)      return query(2 * treeIndex + 1, lo, mid, i);    return query(2 * treeIndex + 2, mid + 1, hi, i);  }}; class Solution { public:  vector<int> treeQueries(int n, vector<vector<int>>& edges,                          vector<vector<int>>& queries) {    LazySegmentTree tree(n);    vector<int> ans;    vector<vector<pair<int, int>>> graph(n + 1);    map<pair<int, int>, int> edgeWeights;     for (const vector<int>& edge : edges) {      const int u = edge[0];      const int v = edge[1];      const int w = edge[2];      graph[u].emplace_back(v, w);      graph[v].emplace_back(u, w);      edgeWeights[{min(u, v), max(u, v)}] = w;    }     // DFS: Euler tour and parent/distance tracking    vector<int> inTime(n + 1);    vector<int> outTime(n + 1);    vector<int> dist(n + 1);    vector<int> parent(n + 1);    int time = 0;     dfs(graph, 1, /*prev=*/-1, time, inTime, outTime, dist, parent);     for (int i = 1; i <= n; ++i)      tree.addRange(inTime[i], inTime[i], dist[i]);     for (const vector<int>& query : queries) {      const int type = query[0];      if (type == 1) {        const int u = query[1];        const int v = query[2];        const int newWeight = query[3];        const auto key = pair<int, int>{min(u, v), max(u, v)};        const int oldWeight = edgeWeights[key];        const int delta = newWeight - oldWeight;        edgeWeights[key] = newWeight;        // Find child node (the one that's not the parent)        const int child = (parent[v] == u) ? v : u;        tree.addRange(inTime[child], outTime[child], delta);      } else {        const int x = query[1];        ans.push_back(tree.query(inTime[x]));      }    }     return ans;  }  private:  void dfs(const vector<vector<pair<int, int>>>& graph, int u, int prev,           int& time, vector<int>& inTime, vector<int>& outTime,           vector<int>& dist, vector<int>& parent) {    inTime[u] = time++;    for (const auto& [v, w] : graph[u]) {      if (v == prev)        continue;      dist[v] = dist[u] + w;      parent[v] = u;      dfs(graph, v, u, time, inTime, outTime, dist, parent);    }    outTime[u] = time - 1;  }};

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