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(Dijkstra 最短路) leetcode 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 0 <= w <= 100.

 

这道题说的就是给了我们一些有向边,又给了一个结点K,问至少需要多少时间才能从K到达任何一个结点。其实是一个求带权的有向图的最短路的的问题,由于带权,所以BFS不能用,我们可以用Dijkstra算法、FLoyd算法以及Bellman-Ford算法。参考链接:http://www.cnblogs.com/grandyang/p/8278115.html

这个题中我用Dijkstra算法,其实,Dijkstra算法有比较固定的模板,本质是利用贪心的。Dijkstra算法由于贪心的性质,所以只能解决权都为正的有向图的最短路的问题。

普通的Dijkstra算法的时间复杂度为O(V2)。

C++代码:

int INF = 0x3f3f3f3f;
class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        int mp[N+1][N+1];
        int dis[N+1];
        int book[N+1];
        for(int i = 1; i <= N; i++){
            for(int j = 1; j <= N; j++){
                mp[i][j] = INF;
            }
        }
        int len = times.size();
        for(int i = 1;i <= len; i++){
            int w = times[i-1][2];
            int u = times[i-1][0];
            int v = times[i-1][1];
            if(w < mp[u][v]){
                mp[u][v] = w;
            }
        }
        for(int i = 1; i <= N; i++){
            dis[i] = mp[K][i];
            book[i] = 0;
        }
        dis[K] = 0;
        book[K] = 1;
        for(int i = 1; i <= N; i++){
            int minn = INF,t = K;
            for(int j = 1; j <= N; j++){
                if(book[j] == 0 && dis[j] < minn){
                    minn = dis[j];
                    t = j;
                }
            }
            book[t] = 1;
            for(int j = 1; j <= N; j++){
                if(book[j] == 0 && dis[t] + mp[t][j] < dis[j] && mp[t][j] < INF){
                    dis[j] = dis[t] + mp[t][j];
                }
            }
        }
        int sum = 0;
        for(int i = 1; i <= N; i++){
            sum = max(sum,dis[i]);
        }
        if(sum == INF)
            return -1;
        return sum;
    }
};

如果是基于优先队列的实现方法的话,时间复杂度就是O(E + VlogV)。

posted @ 2019-05-20 16:05 PJCK 阅读(...) 评论(...) 编辑 收藏