第40天。
今天的题目是Network Delay Time:
一道图的题目,比较常规,用Dijkstra求单源最短路,然后取距离最远的那个即可得到Network Delay Time:
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   | int minDisNode(vector<bool> &visited, vector<int> &dis) {     int min_v = INT_MAX, min_i = -1;;     for(int j = 0;j < dis.size();j++) {         if (!visited[j] && dis[j] < min_v) {             min_v = dis[j];             min_i = j;         }     }     return min_i; } int networkDelayTime(vector<vector<int>>& times, int N, int K) {     if (times.size() == 0 || N==0 || K <= 0) return -1;           vector<vector<int>> graph(N, vector<int>(N, INT_MAX));     for(auto &t: times) {         graph[t[0]-1][t[1]-1] = t[2];     }
      K--;     vector<int> dis(N, INT_MAX);     vector<bool> visited(N, false);     visited[K] = true;     for(int i = 0;i < dis.size(); i++) {         dis[i] = graph[K][i];     }     dis[K] = 0;
      for(int i = 1;i < N; i++) {                  int j = minDisNode(visited, dis);         if (j == -1) return -1;                  visited[j] = true;         for(int k = 0;k < dis.size(); k++) {             if (graph[j][k] != INT_MAX) {                 dis[k] = min(dis[k], dis[j] + graph[j][k]);             }         }     }          int res = 0;     for(int i = 0;i < N;i++) {         if (dis[i] != INT_MAX)             res = max(res, dis[i]);     }     return res; }
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