LeetCode每日一题(1334. Find the City With the Smallest Number of Neighbors at a Threshold Distance)

该博客讨论了一个图论问题，目标是在给定的城市网络中找到一个城市，这个城市的最远邻居不超过特定距离阈值，并且当存在多个这样的城市时，选择邻居数量最多的城市。这个问题涉及到路径距离计算和最优化策略。文章提供了具体的算法实现来解决这个问题，包括Dijkstra算法或者Floyd-Warshall算法的变种。

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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.

计算任意两点之间的距离，然后查找阈值范围内的节点


impl Solution {
    pub fn find_the_city(n: i32, edges: Vec<Vec<i32>>, distance_threshold: i32) -> i32 {
        let mut d = vec![vec![i32::MAX; n as usize]; n as usize];
        for e in edges {
            d[e[0] as usize][e[1] as usize] = e[2];
            d[e[1] as usize][e[0] as usize] = e[2];
        }
        loop {
            let mut updated = false;
            for i in 0..n as usize {
                for j in 0..n as usize {
                    if i == j {
                        continue;
                    }
                    if d[i][j] != i32::MAX {
                        for k in 0..n as usize {
                            if i == k || j == k {
                                continue;
                            }
                            if d[j][k] != i32::MAX && d[i][k] > d[i][j] + d[j][k] {
                                d[i][k] = d[i][j] + d[j][k];
                                d[k][i] = d[i][k];
                                updated = true;
                            }
                        }
                    }
                }
            }
            if !updated {
                break;
            }
        }
        let mut min = i32::MAX;
        let mut ans = 0;
        for i in 0..n as usize {
            let c = d[i].clone().into_iter().filter(|&v| v <= distance_threshold).count() as i32;
            if c <= min {
                min = c;
                ans = i as i32;
            }
        }
        ans
    }
}