Network Delay Time

最新推荐文章于 2018-12-16 01:30:47 发布

caijhBlog

最新推荐文章于 2018-12-16 01:30:47 发布

阅读量271

点赞数

CC 4.0 BY-SA版权

分类专栏：算法文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/caijhBlog/article/details/79123316

算法专栏收录该内容

21 篇文章

订阅专栏

本文探讨了一个关于网络节点信号传播的问题。在一个包含N个节点的网络中，从特定节点K发送信号，如何计算所有节点接收信号的时间？若无法送达则返回-1。采用Bellman-Ford算法解决这一最短路径问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Network Delay Time

A.题意

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:
1.N will be in the range [1, 100].
2.K will be in the range [1, N].
3.The length of times will be in the range [1, 6000].
4.All edges times[i] = (u, v, w) will have 1 <= u, v <= N and 1 <= w <= 100.

B.思路

很明显这道题是最短路问题，这里我用了bellman-ford算法，通过不断更新dist的值，最后可以获得最小值。

C.代码实现

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        int max_Distance = 100 * 100;
        vector<int> dist(N, max_Distance);
        dist[K - 1] = 0;

        for (int i = 1; i < N; i++)
        {
            for (auto time : times)
            {
                int u = time[0] - 1, v = time[1] - 1, w = time[2];
                dist[v] = min(dist[v], dist[u] + w);
            }
        }
        int temp = *max_element(dist.begin(),dist.end());
        return temp == max_Distance ? -1 : temp;
    }
};