algorithm/problem/leetcode/743

743. 网络延迟时间

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有 n 个网络节点，标记为 1 到 n。

给你一个列表 times，表示信号经过 有向 边的传递时间。 times[i] = (ui, vi, wi)，其中 ui 是源节点，vi 是目标节点， wi 是一个信号从源节点传递到目标节点的时间。

现在，从某个节点 K 发出一个信号。需要多久才能使所有节点都收到信号？如果不能使所有节点收到信号，返回 -1 。

示例 1：

输入：times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
输出：2

示例 2：

输入：times = [[1,2,1]], n = 2, k = 1
输出：1

示例 3：

输入：times = [[1,2,1]], n = 2, k = 2
输出：-1

提示：

• 1 <= k <= n <= 100
• 1 <= times.length <= 6000
• times[i].length == 3
• 1 <= ui, vi <= n
• ui != vi
• 0 <= wi <= 100
• 所有 (ui, vi) 对都 互不相同（即，不含重复边）

SPFA算法（链式向前星建图）

class SPFA {
    final int INF = Integer.MAX_VALUE;
    // dists->distances, counts->记录顶点入队次数，用于负圈检测, inQueue->是否在当前队列
    int dists[], counts[], inQueue[];
    // n->E, m->V, s->start, [heads, nexts, tos, weights]->链式向前星
    int n, m, s, heads[], nexts[], tos[], weights[];

    public SPFA(int n, int s, int edges[][]) {
        this.n = n;
        this.s = s;
        m = edges.length;
        int edgeNum = 1;  // range: [2, 2*m+1]
        dists = new int[n+1];
        counts = new int[n+1];
        inQueue = new int[n+1];
        heads = new int[n + 1];
        tos = new int[2 * m + 2];
        nexts = new int[2 * m + 2];
        weights = new int[2 * m + 2];
        for (int[] edge : edges) {  // 链式向前星建图
            int u = edge[0], v = edge[1], w = edge[2];
            // 添加边 (u, v)
            ++edgeNum;
            nexts[edgeNum] = heads[u];
            heads[u] = edgeNum;
            tos[edgeNum] = v;
            weights[edgeNum] = w;
            // 添加反向边 (v, u)
            // ++edgeNum;
            // nexts[edgeNum] = heads[v];
            // heads[v] = edgeNum;
            // tos[edgeNum] = u;
            // weights[edgeNum] = 0;
        }
    }

    public boolean bfs() {
        Arrays.fill(dists, INF);
        Deque<Integer> q = new LinkedList<>();
        q.offer(s);
        dists[s] = 0;
        while (!q.isEmpty()) {
            int p = q.poll();
            inQueue[p] = 0;
            for (int eg = heads[p]; eg != 0; eg = nexts[eg]) {
                int to = tos[eg];
                if (dists[to] > dists[p] + weights[eg]) {
                    dists[to] = dists[p] + weights[eg];
                    if (inQueue[to] == 0) {
                        inQueue[to] = 1;
                        q.offer(to);
                        if (counts[to] > n - 1) {  // 负圈检测
                            System.out.println("Negative Cycle Found!");
                            return false;
                        }
                        else {
                            counts[to] += 1;
                        }
                    }
                }
            }
        }
        return true;
    }
}

class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        int res = 0;
        SPFA spfa = new SPFA(n, k, times);
        spfa.bfs();
        for (int i = 1; i <= n; i++) {  // 找最短路长度最大者
            int dist = spfa.dists[i];
            if (dist == spfa.INF) return -1; // 存在距离未更新的顶点
            if (dist > res) res = dist;
        }
        return res;
    }
}

SPFA算法（邻接表建图）

一些细节也不一样

• 这里数组大小是n，所以index要-1
• 这里直接用队列判定是否在队列中，而不是定义一个数组

class SPFA {
    final int INF = Integer.MAX_VALUE;
    // n->num, s->start, dists->distances, counts->记录顶点入队次数，用于负圈检测
    int n, s, dists[], counts[];  
    List<int[]> g [];
    public SPFA(int n, int s, List<int[]> g[]) {
        this.n = n; this.s = s; this.g = g;
        this.dists = new int[n]; this.counts = new int[n];
    }
    public boolean bfs() {
        Arrays.fill(dists, INF);
        dists[s - 1] = 0;
        Deque<Integer> q = new LinkedList<>();
        q.add(s - 1);
        while (!q.isEmpty()) {
            int u = q.poll();
            for (int vw[] : g[u]) {
                int v = vw[0], w = vw[1];
                if (dists[u] + w < dists[v]) {  // 松弛u的邻接顶点
                    dists[v] = dists[u] + w;
                    if (!q.contains(v)) {  // 入队的前提是此时v不在q中，否则程序虽正确，但复杂度将不再是O(|V||E|)
                        q.add(v);
                        if (counts[v] > n - 1) {  // 负圈检测
                            System.out.println("Negative Cycle Found!");
                            return false;
                        } else {
                            counts[v] += 1;
                        }
                    }
                }
            }
        }
        return true;
    }
}
class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        List<int[]> g[] = new ArrayList[n];
        Arrays.setAll(g, i -> new ArrayList<>());
        for(int[] e : times) {
            int u = e[0]-1, v = e[1]-1, w = e[2];
            g[u].add(new int[]{v, w});
        }
        int res = 0;
        SPFA spfa = new SPFA(n, k, g);
        spfa.bfs();
        for(int dist : spfa.dists){ // 找最短路长度最大者
            if(dist == spfa.INF) return -1; // 存在距离未更新的顶点
            if(dist > res) res = dist;
        }
        return res;
    }
}

Dijkstra算法

class Solution {
    public int networkDelayTime(int[][] times, int n, int k) {
        int g[][] = new int[n][n];
        for (int i = 0; i < n; i++) Arrays.fill(g[i], -1);
        for (int[] time : times) {
            int cur = time[0] - 1, next = time[1] - 1, len = time[2];
            g[cur][next] = len;
        }
        int dis[] = new int[n];
        Arrays.fill(dis, Integer.MAX_VALUE);
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[1] - b[1]);
        dis[k - 1] = 0;
        pq.offer(new int[]{k-1, 0});
        while (!pq.isEmpty()) {
            int poll[] = pq.poll(), cur = poll[0], d = poll[1];
            for (int i = 0; i < n; i++) {
                if (g[cur][i] == -1) continue;
                if (d + g[cur][i] < dis[i]) {
                    dis[i] = d + g[cur][i];
                    pq.offer(new int[]{i, dis[i]});
                }
            }
        }
        int res = Arrays.stream(dis).max().getAsInt();
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}