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.
\n
Example 1:
\n\nInput: n = 5, queries = [[2,4],[0,2],[0,4]]
\n\nOutput: [3,2,1]
\n\nExplanation:
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2024/06/28/image8.jpg\")
After the addition of the road from 2 to 4, the length of the shortest path from 0 to 4 is 3.
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2024/06/28/image9.jpg\")
After the addition of the road from 0 to 2, the length of the shortest path from 0 to 4 is 2.
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2024/06/28/image10.jpg\")
After the addition of the road from 0 to 4, the length of the shortest path from 0 to 4 is 1.
\nExample 2:
\n\nInput: n = 4, queries = [[0,3],[0,2]]
\n\nOutput: [1,1]
\n\nExplanation:
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2024/06/28/image11.jpg\")
After the addition of the road from 0 to 3, the length of the shortest path from 0 to 3 is 1.
\n\n![\"\"](\"https://assets.leetcode.com/uploads/2024/06/28/image12.jpg\")
After the addition of the road from 0 to 2, the length of the shortest path remains 1.
\n\n
Constraints:
\n\n3 <= n <= 5001 <= queries.length <= 500queries[i].length == 20 <= queries[i][0] < queries[i][1] < n1 < queries[i][1] - queries[i][0]Compiled with clang 17 using the latest C++ 23 standard, and libstdc++ provided by GCC 13.
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