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Problem solution · C++

Minimum Weighted Subgraph With the Required Paths II

Minimum Weighted Subgraph With the Required Paths II: a C++ solution using depth-first search. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Depth-first search
Source
walkccc LeetCode Solutions
Length
85 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

Depth-first search

For Minimum Weighted Subgraph With the Required Paths II, the implementation follows one branch at a time, making it suitable for components, trees, backtracking, or dependency exploration.

  1. Define the state carried into one recursive or stack frame.
  2. Mark or choose the current state before exploring children.
  3. Combine child results or undo the choice when the branch finishes.

Code notes

  • 85 lines of C++ from the credited upstream file 3553.cpp.
  • The implementation visibly relies on sequence storage.
  • 7 loop blocks detected, together with recursive traversal.

Complexity

Count unique states for graph traversal; for backtracking, count the branching factor and maximum depth.

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 codeMinimum Weighted Subgraph With the Required Paths II · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  // Similar to 2846. Minimum Edge Weight Equilibrium Queries in a Tree  vector<int> minimumWeight(vector<vector<int>>& edges,                            vector<vector<int>>& queries) {    const int n = edges.size() + 1;    const int m = ceil(log2(n));    vector<int> ans;    vector<vector<pair<int, int>>> graph(n);    // jump[i][j] := the 2^j-th ancestor of i    vector<vector<int>> jump(n, vector<int>(m));    // depth[i] := the depth of i    vector<int> depth(n);    // dist[i] := the distance from root to i    vector<int> dist(n);     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);    }     dfs(graph, 0, /*prev=*/-1, jump, depth, dist);     for (int j = 1; j < m; ++j)      for (int i = 0; i < n; ++i)        jump[i][j] = jump[jump[i][j - 1]][j - 1];     for (const vector<int>& query : queries) {      const int src1 = query[0];      const int src2 = query[1];      const int dest = query[2];      ans.push_back((distance(src1, src2, jump, depth, dist) +                     distance(src1, dest, jump, depth, dist) +                     distance(src2, dest, jump, depth, dist)) /                    2);    }     return ans;  }  private:  void dfs(const vector<vector<pair<int, int>>>& graph, int u, int prev,           vector<vector<int>>& jump, vector<int>& depth, vector<int>& dist) {    for (const auto& [v, w] : graph[u]) {      if (v == prev)        continue;      jump[v][0] = u;      depth[v] = depth[u] + 1;      dist[v] = dist[u] + w;      dfs(graph, v, u, jump, depth, dist);    }  }   // Returns the lca(u, v) by binary jump.  int getLCA(int u, int v, const vector<vector<int>>& jump,             const vector<int>& depth) {    // v is always deeper than u.    if (depth[u] > depth[v])      return getLCA(v, u, jump, depth);    // Jump v to the same height of u.    for (int j = 0; j < jump[0].size(); ++j)      if (depth[v] - depth[u] >> j & 1)        v = jump[v][j];    if (u == v)      return u;    // Jump u and v to the node right below the lca.    for (int j = jump[0].size() - 1; j >= 0; --j)      if (jump[u][j] != jump[v][j]) {        u = jump[u][j];        v = jump[v][j];      }    return jump[u][0];  }   // Returns the distance between u and v.  int distance(int u, int v, const vector<vector<int>>& jump,               const vector<int>& depth, const vector<int>& dist) {    const int lca = getLCA(u, v, jump, depth);    return dist[u] + dist[v] - 2 * dist[lca];  }};

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