This problem is a graph theory problem and can be solved using priority queue and adjacency list. Here are the steps to solve this problem:

1. Parse the input: First, we need to parse the input. We have n nodes and m edges. The edges are given in the form of two arrays a and b, where there is an edge between a[i] and b[i].

2. Create the graph: We need to create an adjacency list to represent the graph. For each edge (a[i], b[i]), we add b[i] to the list of neighbors of a[i] and vice versa.

3. Initialize the priority queue: We initialize a priority queue to keep track of the nodes to visit. The priority queue is ordered in such a way that the smallest node is always at the top. We start by adding the smallest node to the queue.

4. Visit the nodes: While the priority queue is not empty, we pop the smallest node from the queue and add it to the sequence. Then, we add all its unvisited neighbors to the queue. To ensure that we don't visit a node more than once, we keep track of the visited nodes.

5. Return the sequence: Once we have visited all the nodes, we return the sequence of visited nodes.

Here is a Python code snippet that implements these steps:

```python
from heapq import *
def smallest_lexicographical_sequence(n, m, a, b):
graph = [[] for _ in range(n+1)]
for i in range(m):
graph[a[i]].append(b[i])
graph[b[i]].append(a[i])
for i in range(n+1):
graph[i].sort()
visited = [False]*(n+1)
sequence = []
queue = [1]
while queue:
node = heappop(queue)
if not visited[node]:
visited[node] = True
sequence.append(node)
for neighbor in graph[node]:
if not visited[neighbor]:
heappush(queue, neighbor)
return sequence
```

This function takes the number of nodes n, the number of edges m, and the arrays a and b as input, and returns the smallest lexicographical sequence of nodes that can be visited.

Question

This problem is a graph theory problem and can be solved using priority queue and adjacency list. Here are the steps to solve this problem:

1. Parse the input: First, we need to parse the input. We have n nodes and m edges. The edges are given in the form of two arrays a and b, where there is an edge between a[i] and b[i].

2. Create the graph: We need to create an adjacency list to represent the graph. For each edge (a[i], b[i]), we add b[i] to the list of neighbors of a[i] and vice versa.

3. Initialize the priority queue: We initialize a priority queue to keep track of the nodes to visit. The priority queue is ordered in such a way that the smallest node is always at the top. We start by adding the smallest node to the queue.

4. Visit the nodes: While the priority queue is not empty, we pop the smallest node from the queue and add it to the sequence. Then, we add all its unvisited neighbors to the queue. To ensure that we don't visit a node more than once, we keep track of the visited nodes.

5. Return the sequence: Once we have visited all the nodes, we return the sequence of visited nodes.

Here is a Python code snippet that implements these steps:

```python
from heapq import *
def smallest_lexicographical_sequence(n, m, a, b):
    graph = [[] for _ in range(n+1)]
    for i in range(m):
        graph[a[i]].append(b[i])
        graph[b[i]].append(a[i])
    for i in range(n+1):
        graph[i].sort()
    visited = [False]*(n+1)
    sequence = []
    queue = [1]
    while queue:
        node = heappop(queue)
        if not visited[node]:
            visited[node] = True
            sequence.append(node)
            for neighbor in graph[node]:
                if not visited[neighbor]:
                    heappush(queue, neighbor)
    return sequence
```

This function takes the number of nodes n, the number of edges m, and the arrays a and b as input, and returns the smallest lexicographical sequence of nodes that can be visited.

Knowee AI · Accepted Answer

This problem is a graph theory problem and can be solved using priority queue and adjacency list. Here are the steps to solve this problem:

1. Parse the input: First, we need to parse the input. We have n nodes and m edges. The edges are given in the form of two arrays a and b, where there is an edge between a[i] and b[i].

2. Create the graph: We need to create an adjacency list to represent the graph. For each edge (a[i], b[i]), we add b[i] to the list of neighbors of a[i] and vice versa.

3. Initialize the priority queue: We initialize a priority queue to keep track of the nodes to visit. The priority queue is ordered in such a way that the smallest node is always at the top. We start by adding the smallest node to the queue.

4. Visit the nodes: While the priority queue is not empty, we pop the smallest node from the queue and add it to the sequence. Then, we add all its unvisited neighbors to the queue. To ensure that we don't visit a node more than once, we keep track of the visited nodes.

5. Return the sequence: Once we have visited all the nodes, we return the sequence of visited nodes.

Here is a Python code snippet that implements these steps:

```python
from heapq import *
def smallest_lexicographical_sequence(n, m, a, b):
    graph = [[] for _ in range(n+1)]
    for i in range(m):
        graph[a[i]].append(b[i])
        graph[b[i]].append(a[i])
    for i in range(n+1):
        graph[i].sort()
    visited = [False]*(n+1)
    sequence = []
    queue = [1]
    while queue:
        node = heappop(queue)
        if not visited[node]:
            visited[node] = True
            sequence.append(node)
            for neighbor in graph[node]:
                if not visited[neighbor]:
                    heappush(queue, neighbor)
    return sequence
```

This function takes the number of nodes n, the number of edges m, and the arrays a and b as input, and returns the smallest lexicographical sequence of nodes that can be visited.

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