743. Network Delay Time(最短路径)

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.

Solution1: floyd(TLE)

class Solution:
    def networkDelayTime(self, times, N, K):
        """
        :type times: List[List[int]]
        :type N: int
        :type K: int
        :rtype: int
        """
        ans = [[float('inf') for i in range(N)] for _ in range(N)]
        for time in times:
            ans[time[0]-1][time[1]-1] = time[2]
        for i in range(N):
            ans[i][i] = 0
        for k in range(N):
            for i in range(N):
                for j in range(N):
                    ans[i][j] = min(ans[i][j],ans[i][k]+ans[k][j])
        return -1 if float('inf') in ans[K-1] else max(ans[K-1])

Solution2:Bellman-Ford(TLE)

class Solution:
    def networkDelayTime(self, times, N, K):
        """
        :type times: List[List[int]]
        :type N: int
        :type K: int
        :rtype: int
        """
        ans = [float('inf') for _ in range(N)]
        ans[K-1] = 0
        for i in range(N):
            for time in times:
                u,v,w = time[0]-1,time[1]-1,time[2]
                ans[v] = min(ans[v],ans[u] + w)
        return -1 if float('inf') in ans else max(ans)

Solution3:dijkstra

from collections import defaultdict
class Solution:
    def networkDelayTime(self, times, N, K):
        """
        :type times: List[List[int]]
        :type N: int
        :type K: int
        :rtype: int
        """
        K -= 1
        nodes = defaultdict(list)
        for u,v,w in times:
            nodes[u-1].append((v-1,w))
        ans = [float('inf') for _ in range(N)]
        ans[K] = 0
        s = set()
        for _ in range(N):
            smallest = min((d,i) for (i,d) in enumerate(ans) if i not in s)[1]
            for v,w in nodes[smallest]:
                if v not in s and ans[smallest]+w < ans[v]:
                    ans[v] = ans[smallest] + w
            s.add(smallest)
        return -1 if float('inf') in ans else max(ans)

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转载自www.cnblogs.com/bernieloveslife/p/9774659.html