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Problem solution · C++

Minimum Cost of a Path With Special Roads

Minimum Cost of a Path With Special Roads: a C++ solution using heap or priority queue. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Heap or priority queue
Source
walkccc LeetCode Solutions
Length
60 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

Heap or priority queue

For Minimum Cost of a Path With Special Roads, the implementation repeatedly takes the currently best candidate from a heap while inserting newly available choices.

  1. Define the priority key and whether the smallest or largest item should lead.
  2. Push each candidate when it becomes eligible.
  3. Discard stale entries when necessary and process the best live candidate.

Code notes

  • 60 lines of C++ from the credited upstream file 2662.cpp.
  • The implementation visibly relies on sequence storage, work queue.
  • 4 loop blocks detected.

Complexity

Count heap pushes and pops; each normally contributes a logarithmic factor in the heap size.

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 Cost of a Path With Special Roads · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  int minimumCost(vector<int>& start, vector<int>& target,                  vector<vector<int>>& specialRoads) {    return dijkstra(specialRoads, start[0], start[1], target[0], target[1]);  }  private:  int dijkstra(const vector<vector<int>>& specialRoads, int srcX, int srcY,               int dstX, int dstY) {    const int n = specialRoads.size();    // dist[i] := the minimum distance of (srcX, srcY) to    // specialRoads[i](x2, y2)    vector<int> dist(specialRoads.size(), INT_MAX);    using P = pair<int, int>;  // (d, u), where u := the i-th specialRoads    priority_queue<P, vector<P>, greater<>> minHeap;     // (srcX, srcY) -> (x1, y1) to cost -> (x2, y2)    for (int u = 0; u < n; ++u) {      const int x1 = specialRoads[u][0];      const int y1 = specialRoads[u][1];      const int cost = specialRoads[u][4];      const int d = abs(x1 - srcX) + abs(y1 - srcY) + cost;      dist[u] = d;      minHeap.emplace(dist[u], u);    }     while (!minHeap.empty()) {      const auto [d, u] = minHeap.top();      minHeap.pop();      if (d > dist[u])        continue;      const int ux2 = specialRoads[u][2];      const int uy2 = specialRoads[u][3];      for (int v = 0; v < n; ++v) {        if (v == u)          continue;        const int vx1 = specialRoads[v][0];        const int vy1 = specialRoads[v][1];        const int vcost = specialRoads[v][4];        // (ux2, uy2) -> (vx1, vy1) to vcost -> (vx2, vy2)        const int newDist = d + abs(vx1 - ux2) + abs(vy1 - uy2) + vcost;        if (newDist < dist[v]) {          dist[v] = newDist;          minHeap.emplace(dist[v], v);        }      }    }     int ans = abs(dstX - srcX) + abs(dstY - srcY);    for (int u = 0; u < n; ++u) {      const int x2 = specialRoads[u][2];      const int y2 = specialRoads[u][3];      // (srcX, srcY) -> (x2, y2) -> (dstX, dstY).      ans = min(ans, dist[u] + abs(dstX - x2) + abs(dstY - y2));    }    return ans;  }};

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