3243. Shortest Distance After Road Addition Queries I

Problem Link

Description

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You are given an integer n and a 2D integer array queries.

There are n cities numbered from 0 to n - 1. Initially, there is a unidirectional road from city i to city i + 1 for all 0 <= i < n - 1.

queries[i] = [ui, vi] represents the addition of a new unidirectional road from city ui to city vi. After each query, you need to find the length of the shortest path from city 0 to city n - 1.

Return an array answer where for each i in the range [0, queries.length - 1], answer[i] is the length of the shortest path from city 0 to city n - 1 after processing the first i + 1 queries.

 

Example 1:

Input: n = 5, queries = [[2,4],[0,2],[0,4]]

Output: [3,2,1]

Explanation:

After the addition of the road from 2 to 4, the length of the shortest path from 0 to 4 is 3.

After the addition of the road from 0 to 2, the length of the shortest path from 0 to 4 is 2.

After the addition of the road from 0 to 4, the length of the shortest path from 0 to 4 is 1.

Example 2:

Input: n = 4, queries = [[0,3],[0,2]]

Output: [1,1]

Explanation:

After the addition of the road from 0 to 3, the length of the shortest path from 0 to 3 is 1.

After the addition of the road from 0 to 2, the length of the shortest path remains 1.

 

Constraints:

• 3 <= n <= 500
• 1 <= queries.length <= 500
• queries[i].length == 2
• 0 <= queries[i][0] < queries[i][1] < n
• 1 < queries[i][1] - queries[i][0]
• There are no repeated roads among the queries.

Solution

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Python3

class Solution:
    def shortestDistanceAfterQueries(self, n: int, queries: List[List[int]]) -> List[int]:
        graph = defaultdict(set)
 
        for i in range(n - 1):
            graph[i].add(i + 1)
        
        def dijkstra():
            dist = [inf] * n
            dist[0] = 0
            pq = [(0, 0)]
 
            while pq:
                d, node = heappop(pq)
 
                if dist[node] != d: continue
 
                for adj in graph[node]:
                    if d + 1 < dist[adj]:
                        dist[adj] = d + 1
                        heappush(pq, (d + 1, adj))
            
            return dist[n - 1]
        
        res = []
        for x, y in queries:
            graph[x].add(y)
            res.append(dijkstra())
        
        return res