886. 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: 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.

单源最短路径，注意不要重复计算

class Solution:
    def reachableNodes(self, edges, M, N):
        """
        :type edges: List[List[int]]
        :type M: int
        :type N: int
        :rtype: int
        """
        import heapq
        a=[[-1 for _ in range(N)] for _ in range(N)]
        for s,t,n in edges: a[s][t]=a[t][s]=n
        res=0
        pq=[]
        vis=[False for _ in range(N)]
        heapq.heappush(pq, (-M,0))
        
        while pq:
            m,s = heapq.heappop(pq)
            m=-m
            if vis[s]: continue
            vis[s]=True
            res+=1
            for i in range(N):
                if a[s][i]<0: continue
                if m>a[s][i] and not vis[i]: heapq.heappush(pq, (-(m-a[s][i]-1),i))
                # some node between s and i are used
                a[i][s] -= min(m, a[s][i])
                res+=min(m, a[s][i])
        
        return res