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)

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: edges = [[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.

Example 2:

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

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.

題目理解：

給定一個無向圖，從0節點開始可以走M步，問最多能訪問多少個節點。注意這個圖的一部分邊被劃分了，也就是向邊中添加了若干個子節點。

解題思路：

找到每一個原始節點能夠走的最長步數，然後按照這個最大步數訪問它周圍的子節點。為了避免重複，所有訪問過的節點的數目都從子節點數目中減掉。最後加上所有能夠訪問的原始節點數目。

找到每一個原始節點能夠走的最大步數可以使用廣度優先搜尋BFS，這裡的搜尋條件是某一個原始節點的最大步數有變化。

為了降低複雜度，使用一個字典儲存每一個原始節點直接連通的節點。

程式碼如下：

class Solution {
    public int reachableNodes(int[][] edges, int M, int N) {
        Queue<Integer> qu = new LinkedList<Integer>();
        Set<Integer> visited = new HashSet<Integer>();
        Map<Integer, Set<Integer>> map = new HashMap<Integer, Set<Integer>>();
        int res = 0;
        int[][] graph = new int[N][N];
        int[] max = new int[N];
        qu.offer(0);
        max[0] = M;
        for(int[] it : graph) {
        	Arrays.fill(it, -1);
        }
        for(int[] it : edges) {
        	graph[it[0]][it[1]] = it[2];
        	graph[it[1]][it[0]] = it[2];
        	if(!map.containsKey(it[0]))
        		map.put(it[0], new HashSet<Integer>());
        	if(!map.containsKey(it[1]))
        		map.put(it[1], new HashSet<Integer>());
        	map.get(it[0]).add(it[1]);
        	map.get(it[1]).add(it[0]);
        }
        while(!qu.isEmpty()) {
        	int start = qu.poll();
        	for(int end : map.getOrDefault(start, new HashSet<Integer>())) {
        		if(graph[start][end] == -1 || start == end)
    				continue;
    			int remain = max[start] - Math.min(max[start], graph[start][end] + 1);
    			if(remain > max[end]) {//continue to visit other nodes
    				qu.offer(end);
    				max[end] = remain;
    			}
        	}
        }
        for(int i = 0; i < N; i++) {
        	int dis = max[i];
        	if(dis > 0) {
        		visited.add(i);
        		for(int j : map.getOrDefault(i, new HashSet<Integer>())) {
        			if(graph[i][j] < 0)
        				continue;
        			if(dis > graph[i][j])
        				visited.add(j);
        			res += Math.min(dis, graph[i][j]);
        			int temp = Math.max(0, graph[i][j] - dis);
        			graph[i][j] = temp;
        			graph[j][i] = temp;
        		}
        	}
        }
        return res + visited.size();
    }
}