There are N network nodes, labelled 1 to N.

Given times, a list of travel times as directed edges times[i] = (u, v, w), where u is the source node, vis the target node, and w is the time it takes for a signal to travel from source to target.

Now, we send a signal from a certain node K. How long will it take for all nodes to receive the signal? If it is impossible, return -1

Note:

1. N will be in the range [1, 100].
2. K will be in the range [1, N].
3. The length of times will be in the range [1, 6000].
4. All edges times[i] = (u, v, w) will have 1 <= u, v <= N and 1 <= w <= 100.

BFS求解，BFS求K點到各個定點的時間，取最大值即可；同時判斷所有點是否連通，程式如下所示：

class Solution {
    private boolean[] visited;
    private int[] dist;
    private int[][] weight;
    private Set<Integer> set = new HashSet<>();
    public int networkDelayTime(int[][] times, int N, int K) {
        visited = new boolean[N];
        dist = new int[N];
        weight = new int[N][N];
        Arrays.fill(dist, Integer.MAX_VALUE);
        for (int i = 0; i < N; ++ i){
            Arrays.fill(weight[i], -1);
        }
        for (int i = 0; i < times.length; ++ i){
            weight[times[i][0] - 1][times[i][1] - 1] = times[i][2];
        }
        ArrayDeque<Integer> que = new ArrayDeque<>();
        que.offer(K-1);
        dist[K-1] = 0;
        int cnt = 0;
        while (!que.isEmpty()){
            int node = que.poll();
            cnt ++;
            visited[node] = false;
            set.add(node);
            for (int i = 0; i < N; ++ i){
                if (weight[node][i] != -1){
                    if (dist[i] > dist[node] + weight[node][i]){
                        dist[i] = dist[node] + weight[node][i];
                        if (!visited[i]){
                            que.offer(i);
                            visited[i] = true;
                        }
                    }
                }
            }
        }
        int res = 0;
        for (int i = 0; i < N; ++ i){
            res = Math.max(res, dist[i]);
        }
        return set.size() == N ? res:-1;
    }
}