LeetCode #743 - Network Delay Time

最新推荐文章于 2025-01-02 19:21:53 发布

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本文深入探讨了网络信号传播的时间延迟问题，通过分析N个网络节点间的信号传递时间，使用图论中的最短路径算法，解决如何计算从特定节点发送信号到所有节点所需最短时间的问题。文章详细介绍了算法实现过程，包括初始化网络边权重、计算各节点距离及优化路径选择等关键步骤。

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题目描述：

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.

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        int result=0;
        vector<vector<int>> edge(N+1,vector<int>(N+1,-1));
        for(auto time:times) edge[time[0]][time[1]]=time[2];
        vector<int> distance(N+1,INT_MAX);
        queue<int> q;
        q.push(K);
        distance[K]=0;
        while(!q.empty())
        {
            int n=q.size();
            for(int i=0;i<n;i++)
            {
                int u=q.front();
                q.pop();
                for(int v=1;v<=N;v++) // 寻找从u->v的边
                {
                    if(edge[u][v]!=-1)
                    {
                        // 用距离关系来判断是否要访问一个节点，避免重复遍历，而不是用哈希表
                        // 从u->v的路径可以更新distance[v]，那么v应该加入队列中，从而可以更新它的领接点
                        if (distance[v]>distance[u]+edge[u][v]) 
                        {
                            q.push(v);
                            distance[v]=min(distance[v],distance[u]+edge[u][v]);
                        }
                    }
                }
            }
        }
        for (int i=1;i<=N;i++) result=max(result,distance[i]);
        if(result==INT_MAX) return -1;
        else return result;
    }
};