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:

N

will be in the range

[1, 100]

.

K

will be in the range

[1, N]

.

The length of times will be in the range

[1, 6000]

.

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;
}
};