[LeetCode]743. Network Delay Time

文章目录

原题
题解
代码
复杂度分析

原题

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:

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

题解

Dijskra

本题是典型的求解最短路径的问题，我们可以用Dijsktra方法进行求解。

Dijskra原理见：https://www.youtube.com/watch?v=NLp9C7AvJhk&t=524s

维基百科：https://zh.wikipedia.org/zh/%E6%88%B4%E5%85%8B%E6%96%AF%E7%89%B9%E6%8B%89%E7%AE%97%E6%B3%95

代码

typedef pair<int, int> pii;

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        vector<vector<pii>> nodes(N + 1, vector<pii>{});
        for (auto time : times) {
            nodes[time[0]].push_back(make_pair(time[1], time[2]));
        }
        // declare vars
        priority_queue<pii, vector<pii>, greater<pii> > pq; // pii 0 for city index, 1 for cost time
        pq.push(make_pair(K, 0));
        vector<int> time(N + 1, INT_MAX);
        time[K] = 0;
        while (!pq.empty()) {
            pii node = pq.top();
            pq.pop();
            // update the time for the sub-nodes
            for (auto snode : nodes[node.first]) {
                if (time[snode.first] > time[node.first] + snode.second) {
                    time[snode.first] = time[node.first] + snode.second;
                    // add the valid sub-node to priority_queue
                    pq.push(make_pair(snode.first, time[snode.first]));
                }
            }
        }
        int a = -1;
        for (int i = 1; i < time.size(); ++i) {
            //cout<<"i:"<<i<<"time:"<<time[i]<<endl;
            if (a < time[i]) a = time[i];
        }
        return a == INT_MAX ? -1 : a;
    }
};