LeetCode #743 Network Delay Time

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.

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.

解题思路

最重要的就是理解题意，题目的意思是：当信号到达了一个结点时，在下一刻可以同时转发给该结点相邻的所有结点，求传送到每一个结点的最终时间。这样，问题就变成了求解从结点 K 到所有结点的单源最短路问题，然后，从中选择最长的一条路所用的时间即为答案。因此本题使用 Dijkstra 算法进行求解。需要注意的是每次从所有的路径中选择最短路径的结点出来后，要将该结点设置为已访问，避免再重复访问。

C++代码实现

class Solution {public:    int networkDelayTime(vector<vector<int>>& times, int N, int K) {        vector<vector<int> > graph(N + 1, vector<int>(N + 1, 60001));        vector<int> cost(N + 1);        vector<bool> visited(N + 1, false);         for (int i = 0; i < times.size(); ++i) { graph[times[i][0]][times[i][1]] = times[i][2]; }        for (int i = 1; i <= N; ++i) { cost[i] = graph[K][i]; }        cost[K] = 0;        visited[K] = true;        for (int i = 1; i <= N; ++i) {            int minNode = findMinNode(cost, visited);            if (minNode == 0) { break; }            for (int j = 1; j <= N; ++j) {                if (!visited[j] && cost[j] > cost[minNode] + graph[minNode][j]) {                    cost[j] = cost[minNode] + graph[minNode][j];                }            }            visited[minNode] = true;        }        return calSum(cost);    }    int findMinNode(vector<int>& cost, vector<bool>& visited) {        int minIndex = 0, minV = 60001;        for (int i = 1; i < cost.size(); ++i) {            if (!visited[i] && minV > cost[i]) {                 minIndex = i;                 minV = cost[i];            }        }        return minIndex;    }    int calSum(vector<int>& cost) {        int sum = 0;        for (int i = 1; i < cost.size(); ++i) {            if (cost[i] == 60001) { return -1; }            if (sum < cost[i]) { sum = cost[i]; }        }        return sum;    }};