LWC 62：743. Network Delay Time
LWC 62：743. Network Delay Time
LWC 62：743. Network Delay Time
Problem:
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 have 1 <= u, v <= N and 1 <= w <= 100.
思路： 求从某个结点k出发到任意结点的距离，在这些距离中找最大值即可。实际上该题是求解任意两个结点之间的最短路径，先采用Dijkstra算法实现一波。最短路径的相关知识可以参考【挑战程序竞赛系列（11）：2.5最短路径】
Java版：
class Edge{
int from;
int to;
int weight;
Edge(int from, int to, int weight){
this.from = from;
this.to = to;
this.weight = weight;
}
}
class Node implements Comparable<Node>{
int v;
int dist;
Node(int v, int dist){
this.v = v;
this.dist = dist;
}
@Override
public int compareTo(Node that) {
return this.dist - that.dist;
}
}
public int networkDelayTime(int[][] times, int N, int K) {
int n = times.length;
List<Edge>[] graph = new ArrayList[N];
for (int i = 0; i < N; ++i) {
graph[i] = new ArrayList<>();
}
for (int i = 0; i < n; ++i) {
int from = times[i][0];
int to = times[i][1];
int weight = times[i][2];
from --;
to --;
graph[from].add(new Edge(from, to, weight));
}
int ans = 0;
boolean[] vis = new boolean[N];
Queue<Node> queue = new PriorityQueue<>();
queue.offer(new Node(K - 1, 0));
int[] dist = new int[N];
Arrays.fill(dist, 0x3f3f3f3f);
while (!queue.isEmpty()) {
Node now = queue.poll();
for (Edge e : graph[now.v]) {
if (!vis[e.from]) {
Node nxt = new Node(e.to, now.dist + e.weight);
dist[e.to] = Math.min(dist[e.to], nxt.dist);
queue.offer(nxt);
}
}
vis[now.v] = true;
}
for (int i = 0; i < N; ++i) {
if (i == K - 1) continue;
ans = Math.max(ans, dist[i]);
}
return ans >= 0x3f3f3f3f ? -1 : ans;
}当然计算任意结点之间的最短距离可以采用warshallFloyd算法，代码如下：
public int networkDelayTime(int[][] times, int N, int K) {
int INF = 0x3f3f3f3f;
int[][] dist = new int[N][N];
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
dist[i][j] = INF;
}
}
for (int[] time : times) {
int from = time[0];
int to = time[1];
int cost = time[2];
from --;
to --;
dist[from][to] = cost;
}
for (int k = 0; k < N; ++k) {
for (int i = 0; i < N; ++i) {
for (int j = 0; j < N; ++j) {
dist[i][j] = Math.min(dist[i][j], dist[i][k] + dist[k][j]);
}
}
}
int ans = 0;
for (int i = 0; i < N; ++i) {
if (i == K - 1) continue;
ans = Math.max(ans, dist[K - 1][i]);
}
return ans >= INF ? -1 : ans;
}warshallFloyd算法的正确性详见知乎高票回答【Floyd算法为什么把k放在最外层？】
Python版：
def networkDelayTime(self, times, N, K):
"""
:type times: List[List[int]]
:type N: int
:type K: int
:rtype: int
"""
INF = 0x3f3f3f3f
dist = [[INF] * N for _ in range(N)]
for time in times:
u = time[0]
v = time[1]
w = time[2]
dist[u - 1][v - 1] = w
for k in xrange(N):
for i in xrange(N):
for j in xrange(N):
dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j])
ans = 0
for i in range(N):
if (i == K - 1): continue
else:
ans = max(ans, dist[K - 1][i])
return -1 if ans >= INF else ansPython Dijkstra 版：
def networkDelayTime(self, times, N, K):
import heapq
"""
:type times: List[List[int]]
:type N: int
:type K: int
:rtype: int
"""
pq = []
INF = float('inf')
dist = [INF] * N
graph = [[] for _ in range(N)]
for time in times:
graph[time[0] - 1].append((time[1] - 1, time[2]))
heapq.heappush(pq, (0, K - 1))
dist[K - 1] = 0
while len(pq):
cur = heapq.heappop(pq)
for to, cost in graph[cur[1]]:
if (dist[to] > dist[cur[1]] + cost):
dist[to] = cost + dist[cur[1]]
heapq.heappush(pq, (dist[to], to))
ans = 0
for i in range(N):
if i == K - 1:
continue
else:
ans = max(ans, dist[i])
return -1 if ans >= INF else ans知道为什么Dijkstra没有发明warshallFloyd么？因为他的名字是D”ijk”stra而不是D”kij”stra，哈哈！