# 1557. Minimum Number of Vertices to Reach All Nodes
Given a directed acyclic graph, with n vertices numbered from 0 to n-1, and an array edges where edges[i] = [fromi, toi]
represents a directed edge from node from i to node to i. Find the smallest set of vertices from which all nodes in the graph are reachable. It's guaranteed that a unique solution exists. Notice that you can return the vertices in any order.
Example 1:
Input: n = 6, edges = [[0,1],[0,2],[2,5],[3,4],[4,2]]
Output: [0,3]
Explanation: It's not possible to reach all the nodes from a single vertex. From 0 we can reach [0,1,2,5]. From 3 we can reach [3,4,2,5]. So we output [0,3].
Example 2:
Input: n = 5, edges = [[0,1],[2,1],[3,1],[1,4],[2,4]]
Output: [0,2,3]
Explanation: Notice that vertices 0, 3 and 2 are not reachable from any other node, so we must include them. Also any of these vertices can reach nodes 1 and 4.
# Solution
Approach 1: find indegrees, then deal with loops.
# Code (Python)
Approach 1:
class Solution:
def findSmallestSetOfVertices(self, n: int, neighbors: List[List[int]]) -> List[int]:
unreachables = set([i for i in range(n)])
zero_indegree_nodes = set([i for i in range(n)])
edges = {}
for entry in neighbors:
if entry[0] not in edges:
edges[entry[0]] = []
edges[entry[0]].append(entry[1])
zero_indegree_nodes.discard(entry[1])
# step 1: find all vertices with indegree = 0, traverse them and mark up reachable nodes
result = list(zero_indegree_nodes)
for node in zero_indegree_nodes:
self._explore(node, unreachables, edges)
# step 2: the rests are loops that are not connected -- also traverse and mark up
while unreachables:
result.append(min(unreachables))
self._explore(min(unreachables), unreachables, edges)
return sorted(result)
def _explore(self, node, not_seen, edges):
if node not in not_seen:
return
not_seen.remove(node)
if node in edges:
for neighbor in edges[node]:
self._explore(neighbor, not_seen, edges)
# Code (C++)
Approach 1:
Approach 2: