## 743. Network Delay Time

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] = (u`

, where _{i}, v_{i}, w_{i})`u`

is the source node, _{i}`v`

is the target node, and _{i}`w`

is the time it takes for a signal to travel from source to target._{i}

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 = 2Output:2

**Example 2:**

Input:times = [[1,2,1]], n = 2, k = 1Output:1

**Example 3:**

Input:times = [[1,2,1]], n = 2, k = 2Output:-1

**Constraints:**

`1 <= k <= n <= 100`

`1 <= times.length <= 6000`

`times[i].length == 3`

`1 <= u`

_{i}, v_{i}<= n`u`

_{i}!= v_{i}`0 <= w`

_{i}<= 100- All the pairs
`(u`

are_{i}, v_{i})**unique**. (i.e., no multiple edges.)

## Rust Solution

```
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);
}
```

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