Network Delay Time - LeetCode

Network Delay Time - LeetCode

题目：
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.

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考前刷的一道题，正好考试考到了，可惜没有打印出来。
题目会给出一幅图中各条边权值，然后给出一个起点，从起点发出一个信号，经过多长时间能够到达所有点。
这道题就是要求起点到最远点的最短距离，用dijkstra算法就能得出结果。

在代码中我使用了set< pair<int, int> >这样的结构来记录待遍历节点。
pair.first是cost，pair.second是节点的序号。
由于set会自动排序，会按照cost从小到大来排，最大的好处是可以记录下cost与节点的对应关系

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int start) {
        start -= 1;
        int n = N;
        int canNotReach = 99999999;
        vector< vector<int> > g(n, vector<int>(n,canNotReach));
        for (int i = 0; i < times.size(); i++) {
            g[times[i][0]-1][times[i][1]-1] = times[i][2];
        }
        // int n = g.size();
        vector<int> dist(n, canNotReach);
        vector<int> prev(n, -1);
        dist[start] = 0;
        set< pair<int, int> > s;
        s.insert(make_pair(0, start));
        set< pair<int,int> >::iterator it;
        int maxd = 0;
        int count = 0;
        while (!s.empty()) {
            count++;
            int mind = (*s.begin()).first;
            maxd = max(maxd, mind);
            int node = (*s.begin()).second;
            s.erase(s.begin());
            for (int i = 0; i < n; i++) {
                if (mind+g[node][i] < dist[i]) {
                    dist[i] = mind+g[node][i];
                    prev[i] = node;
                    for (it = s.begin(); it != s.end(); it++) {
                        if ((*it).second == i) {
                            s.erase(it);
                            break;
                        }
                    }
                    s.insert(make_pair(dist[i], i));
                }
            }
        }
        if (count < n) return -1;
        return maxd;
    }
};