743. Network Delay Time

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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.

Example 1:

Input: times = [[2,1,1],[2,3,1],[3,4,1]], N = 4, K = 2
Output: 2

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.

class Solution {
public int networkDelayTime(int[][] times, int N, int K) {
    double[][] disTo = new double[N][N];
    for (int i = 0; i < N; i++) {
        Arrays.fill(disTo[i], Double.POSITIVE_INFINITY);
    }
    for (int i = 0; i < N; i++) {
        disTo[i][i] = 0;
    }
    for (int[] edge: times) {
        disTo[edge[0] - 1][edge[1] - 1] = edge[2];
    }
    for (int k = 0; k < N; k++) {
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (disTo[i][j] > disTo[i][k] + disTo[k][j])
                    disTo[i][j] = disTo[i][k] + disTo[k][j];
            }
        }
    }
    double max = Double.MIN_VALUE;
    for (int i = 0; i < N; i++) {
        if (disTo[K - 1][i] == Double.POSITIVE_INFINITY) return -1;//如果有一个节点从k到不了，就说明无法完成任务返回-1
        max = Math.max(max, disTo[K - 1][i]);
    }
    return (int) max;
}
}

1. floyd-warshall

初始化的时候i==j置0，表示没有走动

先把最终矩阵求出来，意思是从i到j的最短路径，

这道题最终转化成从节点k开始，最多要花多少步才能传播到所有节点，所以是上面的矩阵中取大值。

比如从2出发，到1要一步，到4要二步，答案就是2.

743. Network Delay Time

标签：list   cer   class   imp   min   solution   end   get   belle   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12244695.html