743. 网络延迟时间

Title Description

You are given a network of n nodes, labeled from 1 to n. You are also given times, a list of travel times as directed edges times[i] = (ui, vi, wi), where ui is the source node, vi is the target node, and wi is the time it takes for a signal to travel from source to target.

We will send a signal from a given node k. Return the time it takes for all the n nodes to receive the signal. If it is impossible for all the n nodes to receive the signal, return -1.

Example 1:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
Output: 2
Example 2:

Input: times = [[1,2,1]], n = 2, k = 1
Output: 1
Example 3:

Input: times = [[1,2,1]], n = 2, k = 2
Output: -1

Constraints:

• 1 <= k <= n <= 100
• 1 <= times.length <= 6000
• times[i].length == 3
• 1 <= ui, vi <= n
• ui != vi
• 0 <= wi <= 100
• All the pairs (ui, vi) are unique. (i.e., no multiple edges.)

题目描述

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）

解题思路

Dijkstra 算法

• Dijkstra算法一般的表述通常有两种方式，一种用永久和临时标号方式，一种是用OPEN,  CLOSE表的方式，这里均采用永久和临时标号的方式。注意该算法要求图中不存在负权边。

• 求出节点 kk 到其余所有点的最短路，其中的最大值就是答案。
• 若存在从 kk 出发无法到达的点，则返回 -1−1。

代码（今天的还有点没搞懂，就先借用官方大大的，嘻嘻）

var networkDelayTime = function(times, n, k) {
    const INF = Number.MAX_SAFE_INTEGER;
    const g = new Array(n).fill(INF).map(() => new Array(n).fill(INF));
    for (const t of times) {
        const x = t[0] - 1, y = t[1] - 1;
        g[x][y] = t[2];
    }

    const dist = new Array(n).fill(INF);
    dist[k - 1] = 0;
    const used = new Array(n).fill(false);
    for (let i = 0; i < n; ++i) {
        let x = -1;
        for (let y = 0; y < n; ++y) {
            if (!used[y] && (x === -1 || dist[y] < dist[x])) {
                x = y;
            }
        }
        used[x] = true;
        for (let y = 0; y < n; ++y) {
            dist[y] = Math.min(dist[y], dist[x] + g[x][y]);
        }
    }

    let ans = Math.max(...dist);
    return ans === INF ? -1 : ans;
};