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.
Example 1:
Input: n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
Output: 3
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 4 for each city are:
City 0 -> [City 1, City 2]
City 1 -> [City 0, City 2, City 3]
City 2 -> [City 0, City 1, City 3]
City 3 -> [City 1, City 2]
Cities 0 and 3 have 2 neighboring cities at a distanceThreshold = 4, but we have to return city 3 since it has the greatest number.
Example 2:
Input: n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
Output: 0
Explanation: The figure above describes the graph.
The neighboring cities at a distanceThreshold = 2 for each city are:
City 0 -> [City 1]
City 1 -> [City 0, City 4]
City 2 -> [City 3, City 4]
City 3 -> [City 2, City 4]
City 4 -> [City 1, City 2, City 3]
The city 0 has 1 neighboring city at a distanceThreshold = 2.
Constraints:
2 <= n <= 100
1 <= edges.length <= n * (n - 1) / 2
edges[i].length == 3
0 <= fromi < toi < n
1 <= weighti, distanceThreshold <= 10^4
All pairs (fromi, toi) are distinct.
阈值距离内邻居最少的城市。
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。
返回能通过某些路径到达其他城市数目最少、且路径距离 最大 为 distanceThreshold 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。
思路
这道题有多种思路，这里我提供一个最经典的，Floyd算法。这个算法就是为了解决类似本题这种问题的。如果了解这个算法，你就可以直接套用代码解决，如果不了解，建议先学一下这个算法是怎么回事再来看这道题。
复杂度
时间O(n^3)
空间O(n^2)
代码
Java实现
class Solution {
public int findTheCity(int n, int[][] edges, int distanceThreshold) {
int[][] graph = new int[n][n];
// 对角线上赋值为0，因为(i, i)的距离就是0
// 其他点对点的距离还未知
for (int i = 0; i < n; i++) {
Arrays.fill(graph[i], Integer.MAX_VALUE);
graph[i][i] = 0;
}
// 更新其他点对点的距离，注意是双向的
for (int[] e : edges) {
int u = e[0];
int v = e[1];
int weight = e[2];
graph[u][v] = weight;
graph[v][u] = weight;
}
// 使用Floyd-Warshall算法来计算最短路径
// k代表中间某个i和j都能到达的点
for (int k = 0; k < n; k++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (graph[i][k] != Integer.MAX_VALUE && graph[k][j] != Integer.MAX_VALUE) {
graph[i][j] = Math.min(graph[i][j], graph[i][k] + graph[k][j]);
}
}
}
}
int minCount = Integer.MAX_VALUE;
int res = -1;
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (i != j && graph[i][j] <= distanceThreshold) {
count++;
}
}
if (count <= minCount) {
minCount = count;
res = i;
}
}
return res;
}
}