You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.
We will send a signal from a given node k. Return the time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.
Example 1:
Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2 Output: 2
Example 2:
Input: times = [[1,2,1]], n = 2, k = 1 Output: 1
Example 3:
Input: times = [[1,2,1]], n = 2, k = 2 Output: -1
Constraints:
1 <= k <= n <= 1001 <= times.length <= 6000times[i].length == 31 <= ui, vi <= nui != vi0 <= wi <= 100- All the pairs
(ui, vi)are unique. (i.e., no multiple edges.)
struct Solution;
use std::collections::VecDeque;
impl Solution {
fn network_delay_time(times: Vec<Vec<i32>>, n: i32, k: i32) -> i32 {
let n = n as usize;
let k = k as usize - 1;
let mut graph: Vec<Vec<(usize, i32)>> = vec![vec![]; n];
for time in times {
let u = time[0] as usize - 1;
let v = time[1] as usize - 1;
let t = time[2];
graph[u].push((v, t));
}
let mut visited = vec![std::i32::MAX; n];
let mut queue = VecDeque::new();
visited[k] = 0;
queue.push_back(k);
while let Some(u) = queue.pop_front() {
for &(v, t) in &graph[u] {
if t + visited[u] < visited[v] {
visited[v] = t + visited[u];
queue.push_back(v);
}
}
}
let max = visited.into_iter().max().unwrap();
if max == std::i32::MAX {
-1
} else {
max
}
}
}
#[test]
fn test() {
let times = vec_vec_i32![[2, 1, 1], [2, 3, 1], [3, 4, 1]];
let n = 4;
let k = 2;
let res = 2;
assert_eq!(Solution::network_delay_time(times, n, k), res);
}