743. Network Delay Time

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, v is 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.

 

Approach #1: C++.[Bellman-Ford]

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        const int MAX_TIME = 101 * 100;
        vector<int> dist(N, MAX_TIME);
        dist[K-1] = 0;
        for (int i = 0; i < N; ++i) {
            for (auto time : times) {
                int u = time[0]-1, v = time[1] - 1, w = time[2];
                dist[v] = min(dist[v], dist[u] + w);
            }
        }
        int temp = *max_element(dist.begin(), dist.end());
        return temp == MAX_TIME ? -1 : temp;
    }
};

　　

Approach #2: C++. [Floyd-Warshall]

// Author: Huahua
// Runtime: 89 ms
class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        constexpr int MAX_TIME = 101 * 100;
        vector<vector<int>> d(N, vector<int>(N, MAX_TIME));
        
        for (auto time : times)   
            d[time[0] - 1][time[1] - 1] = time[2];
        
        for (int i = 0; i < N; ++i)
            d[i][i] = 0;
        
        for (int k = 0; k < N; ++k)
            for (int i = 0; i < N; ++i)
                for (int j = 0; j < N; ++j)                    
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
        
        int max_time = *max_element(d[K - 1].begin(), d[K - 1].end());        
        return max_time >= MAX_TIME ? - 1 : max_time;
    }
};

　　

Appraoch #3: Java. [Dijkstra's Algorithm]

class Solution {
    Map<Integer, Integer> dist;
    public int networkDelayTime(int[][] times, int N, int K) {
        Map<Integer, List<int[]>> graph = new HashMap();
        for (int[] edge : times) {
            if (!graph.containsKey(edge[0])) {
                graph.put(edge[0], new ArrayList<int[]>());
            }
            graph.get(edge[0]).add(new int[]{edge[1], edge[2]});
        }
        dist = new HashMap();
        for (int node = 1; node <= N; ++node) {
            dist.put(node, Integer.MAX_VALUE);
        }
        dist.put(K, 0);
        boolean[] seen = new boolean[N+1];
        while (true) {
            int candNode = -1;
            int candDist = Integer.MAX_VALUE;
            for (int i = 1; i <= N; ++i) {
                if (!seen[i] && dist.get(i) < candDist) {
                    candDist = dist.get(i);
                    candNode = i;
                }
            }
            if (candNode < 0) break;
            seen[candNode] = true;
            if (graph.containsKey(candNode)) {
                for (int[] info : graph.get(candNode))
                    dist.put(info[0], Math.min(dist.get(info[0]), dist.get(candNode) + info[1]));
            }
        }
        int ans = 0;
        for (int cand : dist.values()) {
            if (cand == Integer.MAX_VALUE) return -1;
            ans = Math.max(ans, cand);
        }
        return ans;
    }
}

　　

 

Appraoch #4: Python[DFS]

class Solution(object):
    def networkDelayTime(self, times, N, K):
        """
        :type times: List[List[int]]
        :type N: int
        :type K: int
        :rtype: int
        """
        graph = collections.defaultdict(list)
        
        for u, v, w in times:
            graph[u].append((w, v))
            
        dist = {node: float('inf') for node in xrange(1, N+1)}
        
        def dfs(node, elapsed):
            if elapsed >= dist[node]: return
            dist[node] = elapsed
            for time, nei in sorted(graph[node]):
                dfs(nei, elapsed + time)
            
        dfs(K, 0)
        
        ans = max(dist.values())
        return ans if ans < float('inf') else -1