# 743. Network Delay Time

## 1. Question

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


## 2. Constraints

• 1 <= k <= n <= 100
• 1 <= times.length <= 6000
• times[i].length == 3
• 1 <= ui, vi <= n
• ui != vi
• 0 <= wi <= 100
• All the pairs (ui, vi) are unique. (i.e., no multiple edges.)

## 4. Solutions

1. Folyd

class Solution {
public int networkDelayTime(int[][] times, int n, int k) {
final int MAX = 999999;

int[][] res = new int[n + 1][n + 1];
for (int[] re : res) {
Arrays.fill(re, MAX);
}
for (int i = 1; i <= n; i++) {
res[i][i] = 0;
}

for (int i = 0; i < times.length; i++) {
res[times[i]][times[i]] = times[i];
}

for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
for (int l = 1; l <= n; l++) {
if (res[j][l] > res[j][i] + res[i][l]) {
res[j][l] = res[j][i] + res[i][l];
}
}
}
}

int max = -1;

for (int i = 1; i <= n; i++) {
if (k == i) {
continue;
}
if (MAX == res[k][i]) {
max = -1;
break;
}
if (res[k][i] > max) {
max = res[k][i];
}
}

return max;
}
}