You are given a directed, weighted graph with n nodes labeled from 0 to n - 1, and an array edges where edges[i] = [ui, vi, wi] represents a directed edge from node ui to node vi with cost wi.
Each node ui has a switch that can be used at most once: when you arrive at ui and have not yet used its switch, you may activate it on one of its incoming edges vi → ui reverse that edge to ui → vi and immediately traverse it.
The reversal is only valid for that single move, and using a reversed edge costs 2 * wi.
Return the minimum total cost to travel from node 0 to node n - 1. If it is not possible, return -1.
\n
Example 1:
\n\nInput: n = 4, edges = [[0,1,3],[3,1,1],[2,3,4],[0,2,2]]
\n\nOutput: 5
\n\nExplanation:
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2025/05/07/e1drawio.png\")
0 → 1 (cost 3).3 → 1 into 1 → 3 and traverse it at cost 2 * 1 = 2.3 + 2 = 5.Example 2:
\n\nInput: n = 4, edges = [[0,2,1],[2,1,1],[1,3,1],[2,3,3]]
\n\nOutput: 3
\n\nExplanation:
\n\n0 → 2 (cost 1), then 2 → 1 (cost 1), then 1 → 3 (cost 1).1 + 1 + 1 = 3.\n
Constraints:
\n\n2 <= n <= 5 * 1041 <= edges.length <= 105edges[i] = [ui, vi, wi]0 <= ui, vi <= n - 11 <= wi <= 1000给你一个包含 n 个节点的有向带权图，节点编号从 0 到 n - 1。同时给你一个数组 edges，其中 edges[i] = [ui, vi, wi] 表示一条从节点 ui 到节点 vi 的有向边，其成本为 wi。
每个节点 ui 都有一个 最多可使用一次 的开关：当你到达 ui 且尚未使用其开关时，你可以对其一条入边 vi → ui 激活开关，将该边反转为 ui → vi 并 立即 穿过它。
反转仅对那一次移动有效，使用反转边的成本为 2 * wi。
返回从节点 0 到达节点 n - 1 的 最小 总成本。如果无法到达，则返回 -1。
\n\n
示例 1:
\n\n输入: n = 4, edges = [[0,1,3],[3,1,1],[2,3,4],[0,2,2]]
\n\n输出: 5
\n\n解释:
\n\n
0 → 1 (成本 3)。3 → 1 反转为 1 → 3 并穿过它，成本为 2 * 1 = 2。3 + 2 = 5。示例 2:
\n\n输入: n = 4, edges = [[0,2,1],[2,1,1],[1,3,1],[2,3,3]]
\n\n输出: 3
\n\n解释:
\n\n0 → 2 (成本 1)，然后 2 → 1 (成本 1)，再然后 1 → 3 (成本 1)。1 + 1 + 1 = 3。\n\n
提示:
\n\n2 <= n <= 5 * 1041 <= edges.length <= 105edges[i] = [ui, vi, wi]0 <= ui, vi <= n - 11 <= wi <= 1000{u, v, w} -> {v, u, 2 * w}, and use Dijkstra."
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