[Swift]LeetCode882. 细分图中的可到达结点 | Reachable Nodes In Subdivided Graph

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Starting with an undirected graph (the "original graph") with nodes from 0 to N-1, subdivisions are made to some of the edges.

The graph is given as follows: edges[k] is a list of integer pairs (i, j, n) such that (i, j) is an edge of the original graph,

and n is the total number of new nodes on that edge. 

Then, the edge (i, j) is deleted from the original graph, n new nodes (x_1, x_2, ..., x_n) are added to the original graph,

and n+1 new edges (i, x_1), (x_1, x_2), (x_2, x_3), ..., (x_{n-1}, x_n), (x_n, j) are added to the original graph.

Now, you start at node 0 from the original graph, and in each move, you travel along one edge. 

Return how many nodes you can reach in at most M moves. 

Example 1:

Input: = [[0,1,10],[0,2,1],[1,2,2]], M = 6, N = 3 Output: 13 Explanation: The nodes that are reachable in the final graph after M = 6 moves are indicated below.  edges

Example 2:

Input: = [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], M = 10, N = 4 Output: 23 edges

Note:

1. 0 <= edges.length <= 10000
2. 0 <= edges[i][0] < edges[i][1] < N
3. There does not exist any i != j for which edges[i][0] == edges[j][0] and edges[i][1] == edges[j][1].
4. The original graph has no parallel edges.
5. 0 <= edges[i][2] <= 10000
6. 0 <= M <= 10^9
7. 1 <= N <= 3000
8. A reachable node is a node that can be travelled to using at most M moves starting from node 0.

---

从具备 0 到 N-1 的结点的无向图（“原始图”）开始，对一些边进行细分。

该图给出以下：edges[k] 是整数对 (i, j, n) 组成的列表，使 (i, j) 是原始图的边。

n 是该边上新结点的总数

而后，将边 (i, j) 从原始图中删除，将 n 个新结点 (x_1, x_2, ..., x_n) 添加到原始图中，

将 n+1 条新边 (i, x_1), (x_1, x_2), (x_2, x_3), ..., (x_{n-1}, x_n), (x_n, j) 添加到原始图中。

如今，你将从原始图中的结点 0 处出发，而且每次移动，你都将沿着一条边行进。

返回最多 M 次移动能够达到的结点数。 

示例 1：

输入：= [[0,1,10],[0,2,1],[1,2,2]], M = 6, N = 3
输出：13
解释：
在 M = 6 次移动以后在最终图中可到达的结点以下所示。
edges

示例 2：

输入：= [[0,1,4],[1,2,6],[0,2,8],[1,3,1]], M = 10, N = 4
输出：23 edges

提示：

1. 0 <= edges.length <= 10000
2. 0 <= edges[i][0] < edges[i][1] < N
3. 不存在任何 i != j 状况下 edges[i][0] == edges[j][0] 且 edges[i][1] == edges[j][1].
4. 原始图没有平行的边。
5. 0 <= edges[i][2] <= 10000
6. 0 <= M <= 10^9
7. 1 <= N <= 3000

---

Runtime: 572 ms

Memory Usage: 21.3 MB

 1 class Solution {
 2     func reachableNodes(_ edges: [[Int]], _ M: Int, _ N: Int) -> Int {
 3         var steps:[Int] = [Int](repeating:-1,count:N)
 4         steps[0] = M
 5         for _ in 0...N
 6         {
 7             var stable:Bool = true
 8             for edge in edges
 9             {
10                 if 0 < steps[edge[0]]
11                 {
12                     let diff:Int = steps[edge[0]] - edge[2] - 1
13                     if steps[edge[1]] < diff
14                     {
15                         steps[edge[1]] = diff
16                         stable = false
17                     }
18                 }
19                 if 0 < steps[edge[1]]
20                 {
21                     let diff:Int = steps[edge[1]] - edge[2] - 1
22                     if steps[edge[0]] < diff
23                     {
24                         steps[edge[0]] = diff
25                         stable = false
26                     }
27                 }
28             }
29             if stable {break}
30         }
31         var res:Int = 0
32         for i in steps
33         {
34             if 0 <= i
35             {
36                 res += 1
37             }
38         }
39         for edge in edges
40         {
41             var cnt:Int = 0
42             if 0 < steps[edge[0]]
43             {
44                 cnt += steps[edge[0]]
45             }
46             if 0 < steps[edge[1]]
47             {
48                 cnt += steps[edge[1]]
49             }
50             res += edge[2] >= cnt ? cnt : edge[2]
51         }
52         return res
53     }
54 }