![\"\"](\"https://assets.leetcode.com/uploads/2024/08/23/problem1334example1.png\")
There are n cities numbered from 0 to n-1. Given the array edges where edges[i] = [fromi, toi, weighti] represents a bidirectional and weighted edge between cities fromi and toi, and given the integer distanceThreshold.
Return the city with the smallest number of cities that are reachable through some path and whose distance is at most distanceThreshold, If there are multiple such cities, return the city with the greatest number.
Notice that the distance of a path connecting cities i and j is equal to the sum of the edges' weights along that path.
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Example 1:
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\nInput: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4\nOutput: 3\nExplanation: The figure above describes the graph. \nThe neighboring cities at a distanceThreshold = 4 for each city are:\nCity 0 -> [City 1, City 2] \nCity 1 -> [City 0, City 2, City 3] \nCity 2 -> [City 0, City 1, City 3] \nCity 3 -> [City 1, City 2] \nCities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.\n\n\n
Example 2:
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2024/08/23/problem1334example0.png\")
\nInput: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2\nOutput: 0\nExplanation: The figure above describes the graph. \nThe neighboring cities at a distanceThreshold = 2 for each city are:\nCity 0 -> [City 1] \nCity 1 -> [City 0, City 4] \nCity 2 -> [City 3, City 4] \nCity 3 -> [City 2, City 4]\nCity 4 -> [City 1, City 2, City 3] \nThe city 0 has 1 neighboring city at a distanceThreshold = 2.\n\n\n
\n
Constraints:
\n\n2 <= n <= 1001 <= edges.length <= n * (n - 1) / 2edges[i].length == 30 <= fromi < toi < n1 <= weighti, distanceThreshold <= 10^4(fromi, toi) are distinct.有 n 个城市，按从 0 到 n-1 编号。给你一个边数组 edges，其中 edges[i] = [fromi, toi, weighti] 代表 fromi 和 toi 两个城市之间的双向加权边，距离阈值是一个整数 distanceThreshold。
返回在路径距离限制为 distanceThreshold 以内可到达城市最少的城市。如果有多个这样的城市，则返回编号最大的城市。
注意，连接城市 i 和 j 的路径的距离等于沿该路径的所有边的权重之和。
\n\n\n\n
示例 1：
\n\n
\n输入：n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4\n输出：3\n解释：城市分布图如上。\n每个城市阈值距离 distanceThreshold = 4 内的邻居城市分别是：\n城市 0 -> [城市 1, 城市 2] \n城市 1 -> [城市 0, 城市 2, 城市 3] \n城市 2 -> [城市 0, 城市 1, 城市 3] \n城市 3 -> [城市 1, 城市 2] \n城市 0 和 3 在阈值距离 4 以内都有 2 个邻居城市，但是我们必须返回城市 3，因为它的编号最大。\n\n\n
示例 2：
\n\n
\n输入：n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2\n输出：0\n解释：城市分布图如上。 \n每个城市阈值距离 distanceThreshold = 2 内的邻居城市分别是：\n城市 0 -> [城市 1] \n城市 1 -> [城市 0, 城市 4] \n城市 2 -> [城市 3, 城市 4] \n城市 3 -> [城市 2, 城市 4]\n城市 4 -> [城市 1, 城市 2, 城市 3] \n城市 0 在阈值距离 2 以内只有 1 个邻居城市。\n\n\n
\n\n
提示：
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