2662. Minimum Cost of a Path With Special Roads

```java
class Solution {
  class Step {
    int x;
    int y;
    int cost;
    Step(int x, int y, int cost) {
      this.x = x;
      this.y = y;
      this.cost = cost;
    }
  }

  public int minimumCost(int[] start, int[] target, int[][] specialRoads) {
    Queue<Step> queue = new PriorityQueue<>((a, b) -> a.cost - b.cost);
    queue.offer(new Step(start[0], start[1], 0)); // put start

    Set<Pair<Integer, Integer>> visited = new HashSet<>();

    while (!queue.isEmpty()) {
      Step cur = queue.poll();
      if (cur.x == target[0] && cur.y == target[1]) {
        return cur.cost;
      }

      Pair position = new Pair(cur.x, cur.y);
      if (visited.contains(position)) {
        continue;
      }
      // key point -> put all kinds of dist to queue, visited record this ending's value
      // here every time, we'll update a smaller value in PQ (poll smaller first)
      visited.add(position); // dist: current to target
      queue.offer(new Step(target[0], target[1], getDist(cur.x, cur.y, target[0], target[1]) + cur.cost));

      for (int[] sp : specialRoads) { // dist: current to sp_end
        queue.offer(new Step(sp[2], sp[3], getDist(cur.x, cur.y, sp[0], sp[1]) + cur.cost + sp[4]));
      }
    }
    return -1;
  }

  private int getDist(int x1, int y1, int x2, int y2) {
    return Math.abs(x1 - x2) + Math.abs(y1 - y2);
  }
}
```