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:
N
will be in the range[1, 100]
.K
will be in the range[1, N]
.- The length of
times
will be in the range[1, 6000]
. - All edges
times[i] = (u, v, w)
will have1 <= u, v <= N
and1 <= w <= 100
.
这道题给了我们一些有向边,又给了一个结点K,问至少需要多少时间才能从K到达任何一个结点。这实际上是一个有向图求最短路径的问题,我们求出K点到每一个点到最短路径,然后取其中最大的一个就是需要的时间了。
可以想成从结点K开始有水流向周围扩散,当水流到达最远的一个结点时,那么其他所有的结点一定已经流过水了。最短路径的常用解法有迪杰克斯特拉算法Dijkstra Algorithm, 弗洛伊德算法Floyd-Warshall Algorithm, 和贝尔曼福特算法Bellman-Ford Algorithm,其中,Floyd算法是多源最短路径,即求任意点到任意点到最短路径,而Dijkstra算法和Bellman-Ford算法是单源最短路径,即单个点到任意点到最短路径。这里因为起点只有一个K,所以使用单源最短路径就行了。这三种算法还有一点不同,就是Dijkstra算法处理有向权重图时,权重必须为正,而另外两种可以处理负权重有向图,但是不能出现负环,所谓负环,就是权重均为负的环。为啥呢,这里要先引入松弛操作Relaxtion,这是这三个算法的核心思想,当有对边 (u, v) 是结点u到结点v,如果 dist(v) > dist(u) + w(u, v),那么 dist(v) 就可以被更新,这是所有这些的算法的核心操作。Dijkstra算法是以起点为中心,向外层层扩展,直到扩展到终点为止。根据这特性,用BFS来实现时再好不过了。
class Solution(object): def networkDelayTime(self, times, N, K): """ :type times: List[List[int]] :type N: int :type K: int :rtype: int """ inf = float('inf') dp = [inf]*(N+1) dp[K] = 0 for i in range(1, N): # N+1 is not neccessary for u,v,w in times: #if v == i: dp[v] = min(dp[v], dp[u]+w) max_d = max(dp[1:]) return -1 if max_d == inf else max_d
注意:(1) 我以为是要加一个if判断,实际上是错的。迭代是针对图里所有边,第一次迭代找到的是source点走一次直接连接的最短路。第二次是走二次的最短路迭代。一直到第n-1次走的最短路。
(2)为啥是1,N而不是N+1。
为什么要循环n-1次?图有n个点,又不能有回路,所以最短路径最多n-1边。又因为每次循环,至少relax一边所以最多n-1次就行了!
dijkstra解法:
class Solution(object): def networkDelayTime(self, times, N, K): """ :type times: List[List[int]] :type N: int :type K: int :rtype: int """ return self.dijkstra(times, N, K) def dijkstra(sefl, times, N, k): G = collections.defaultdict(dict) for n1, n2, w in times: G[n1][n2] = w inf = float('inf') dist = {node:inf for node in range(1, N+1)} dist[k] = 0 nodes = set(range(1, N+1)) while nodes: node = min(nodes, key=dist.get) # update dist for n in G[node]: dist[n] = min(dist[n], dist[node] + G[node][n]) # node visited nodes.remove(node) max_dist = -1 for node in dist: if node != k: max_dist = max(max_dist, dist[node]) return max_dist if max_dist != inf else -1
或者将dist数据结构修改为list
class Solution(object): def networkDelayTime(self, times, N, K): """ :type times: List[List[int]] :type N: int :type K: int :rtype: int """ return self.dijkstra(times, N, K) def dijkstra(sefl, times, N, k): G = collections.defaultdict(dict) for n1, n2, w in times: G[n1][n2] = w inf = float('inf') dist = [inf]*(N+1) dist[k] = 0 nodes = set(range(1, N+1)) while nodes: node = min(nodes, key=dist.__getitem__) # update dist for n in G[node]: dist[n] = min(dist[n], dist[node] + G[node][n]) # node visited nodes.remove(node) max_dist = max(dist[1:]) return max_dist if max_dist != inf else -1