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 u_i is the source node, v_i is the target node, and w_i 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 (u_i, v_i) are unique. (i.e., no multiple edges.)

3. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/network-delay-time 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

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][0]][times[i][1]] = times[i][2];
           }
   
           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;
       }
   }