https://leetcode-cn.com/problems/network-delay-time/submissions/

// n <= 100
class Solution {
    int N = 105, M = 6005;
    // （邻接表-链式前向星）
    int[] w = new int[M];       // 边的权重
    int[] edge = new int[M];    // 边指向的节点
    int[] head = new int[N];    // 存储的是某个节点的链表（节点指向边的结合）的第一个节点
    int[] next = new int[M];    // 表示链表中的下一条边

    int[] dist = new int[N];    // 起点到i的最短路
    boolean[] vis = new boolean[N];

    int n, k;
    int idx = 0;
    int inf = 0x3f3f3f3f;

    void add(int a, int b, int c) { 
        // 链表中添加新节点
        w[idx] = c;
        edge[idx] = b;
        next[idx] = head[a];

        head[a] = idx;
        idx++;
    }

    void dijkstra() {
        // 起始先将所有的点标记为「未更新」和「距离为正无穷」
        Arrays.fill(dist, inf);
        Arrays.fill(vis, false);
        // 只有起点最短距离为 0
        dist[k] = 0;
        // (点编号， 到起点的距离)
        PriorityQueue<int[]> pq = new PriorityQueue<>((x, y) -> {
            return x[1] - y[1];
        });
        pq.offer(new int[]{k, 0});
        while (!pq.isEmpty()) {
            int[] poll = pq.poll();
            int index = poll[0], distance = poll[1];
            if (vis[index]) continue;
            vis[index] = true;
            // 标记该点「已更新」，并使用该点更新其他点的「最短距离」
            for (int i = head[index]; i != -1; i = next[i]) {   // 链表最末端都是-1
                int to = edge[i];
                if (dist[to] > dist[index] + w[i]) {
                    dist[to] = dist[index] + w[i];
                    pq.offer(new int[]{to, dist[to]});
                }
            }
        }
    }

    public int networkDelayTime(int[][] times, int n, int k) {
        this.n = n;
        this.k = k;
        // 初始化每个节点的链表的头结点
        Arrays.fill(head, -1);
        // 建图
        for (int[] time: times) {
            int from = time[0], to = time[1], val = time[2];
            add(from, to, val);
        }
        // 最短路
        dijkstra();
        // 遍历答案
        int ans = 0;
        for (int i = 1; i <= n; i++) {
            ans = Math.max(ans, dist[i]);
        }
        return ans == inf? -1: ans;
    }
}

进阶：https://leetcode-cn.com/problems/count-nodes-with-the-highest-score/solution/gong-shui-san-xie-jian-tu-dfs-by-ac_oier-ujfo/

// 链式前向星 求 每一个节点能到达的节点数量
class Solution {
    int[] cnt;
    int N = 100010, M = 100010 * 2;
    int[] head = new int[N];
    int[] next = new int[M];
    int[] edge = new int[M];

    int idx = 0;
    void add(int x, int y) {
        edge[idx] = y;
        next[idx] = head[x];
        head[x] = idx;
        idx++;
    }

    public int countHighestScoreNodes(int[] parents) {
        int n = parents.length;
        cnt = new int[n]; 
        Arrays.fill(head, -1);

        for (int i = 0; i < n; i++) {
            int from = parents[i];
            int to = i;
            if (from == -1) continue;
            add(from, to);
        }

        // dfs
        dfs(0);
        long maxVal = 0L;
        int maxNum = 0;
        for (int i = 0; i < n; i++) {
            long tmp = 0;
            if (parents[i] == -1) { // 根节点
                long res = 1;
                for (int j = head[i]; j != -1; j = next[j]) {
                    int to = edge[j];
                    res *= cnt[to];
                }
                tmp = res;
            } else if (head[i] == -1) {  // 叶子节点
                tmp = n-1;
            } else {    // 非叶子节点
                long res = 1;
                for (int j = head[i]; j != -1; j = next[j]) {
                    int to = edge[j];
                    res *= cnt[to];
                }
                tmp = res * (n-cnt[i]);
            }
            if (tmp > maxVal) {
                maxVal = tmp; 
                maxNum = 1;
            } else if (tmp == maxVal) {
                maxNum++;
            }
        }
        return maxNum;
    }
    
    // 求以u为根节点的子树的节点数量
    public int dfs(int u) {
        int num = 1;
        // 遍历所有的边
        for (int i = head[u]; i != -1; i = next[i]) {
            int to = edge[i];
            num += dfs(to);
        }
        cnt[u] = num;
        return num;
    }
}