leetcode 743. Network Delay Time

743. Network Delay Time

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.

1、BFS。如果到一个点，发现接下去要找的点到达的时间比之前到过的时间要短，才push进queue。不然没有意义。

2、map存到达时间。也可以当做bfs的hash表使用。

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K)
    {
        int ret = INT_MIN;
        map<int, vector<pair<int, int> > > route;   //pair 存的是 终点-时间
        for (int i = 0; i < times.size(); i ++)
        {
            route[times[i][0]].push_back(make_pair(times[i][1], times[i][2]));
        }
        map<int, int> arrive_time;  //到达i点的最快时间。也可以当做hash表存已经遍历过的点
        arrive_time[K] = 0;
        queue<int> que;
        que.push(K);
        while (!que.empty())
        {
            int top = que.front();
            que.pop();
            for (int i = 0; i < route[top].size(); i++)
            {
                if (arrive_time.find(route[top][i].first) == arrive_time.end() ||
                    arrive_time[route[top][i].first] > arrive_time[top] + route[top][i].second
                    ) //第一次遍历 或 当前这个路径的时间更少，那就更新
                {
                    arrive_time[route[top][i].first] = arrive_time[top] + route[top][i].second;
                    que.push(route[top][i].first);
                }
            }
        }
        if (arrive_time.size() != N)
            return -1;
        for (auto it : arrive_time)
            ret = max(ret, it.second);
        return ret;
    }
};