花花酱 LeetCode 2662. Minimum Cost of a Path With Special Roads

You are given an array start where start = [startX, startY] represents your initial position (startX, startY) in a 2D space. You are also given the array target where target = [targetX, targetY] represents your target position (targetX, targetY).

The cost of going from a position (x1, y1) to any other position in the space (x2, y2) is |x2 - x1| + |y2 - y1|.

There are also some special roads. You are given a 2D array specialRoads where specialRoads[i] = [x1i, y1i, x2i, y2i, costi] indicates that the ith special road can take you from (x1i, y1i) to (x2i, y2i) with a cost equal to costi. You can use each special road any number of times.

Return the minimum cost required to go from (startX, startY) to (targetX, targetY).

Example 1:

Input: start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]
Output: 5
Explanation: The optimal path from (1,1) to (4,5) is the following:
- (1,1) -> (1,2). This move has a cost of |1 - 1| + |2 - 1| = 1.
- (1,2) -> (3,3). This move uses the first special edge, the cost is 2.
- (3,3) -> (3,4). This move has a cost of |3 - 3| + |4 - 3| = 1.
- (3,4) -> (4,5). This move uses the second special edge, the cost is 1.
So the total cost is 1 + 2 + 1 + 1 = 5.
It can be shown that we cannot achieve a smaller total cost than 5.

Example 2:

Input: start = [3,2], target = [5,7], specialRoads = [[3,2,3,4,4],[3,3,5,5,5],[3,4,5,6,6]]
Output: 7
Explanation: It is optimal to not use any special edges and go directly from the starting to the ending position with a cost |5 - 3| + |7 - 2| = 7.

Constraints:

• start.length == target.length == 2
• 1 <= startX <= targetX <= 105
• 1 <= startY <= targetY <= 105
• 1 <= specialRoads.length <= 200
• specialRoads[i].length == 5
• startX <= x1i, x2i <= targetX
• startY <= y1i, y2i <= targetY
• 1 <= costi <= 105

Solution: Dijkstra

1. Create a node for each point in special edges as well as start and target.
2. Add edges for special roads (note it’s directional)
3. Add edges for each pair of node with default cost, i.e. |x1-x2| + |y1-y2|
4. Run Dijkstra’s algorithm

Time complexity: O(n²logn)
Space complexity: O(n²)