3244. Shortest Distance After Road Addition Queries II

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3244. Shortest Distance After Road Addition Queries II

Description

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.

There are no two queries such that queries[i][0] < queries[j][0] < queries[i][1] < queries[j][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:

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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:

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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 <= 10^5
  • 1 <= queries.length <= 10^5
  • 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.
  • There are no two queries such that i != j and queries[i][0] < queries[j][0] < queries[i][1] < queries[j][1].

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<int> nodes;

    vector<int> shortestDistanceAfterQueries(int n,
                                             vector<vector<int>>& queries) {
        nodes.resize(n - 1, 0);
        iota(nodes.begin(), nodes.end(), 0);
        vector<int> res;
        int count = n - 1;
        for (auto q : queries) {
            int l = find(q[0]), r = find(q[1] - 1);
            // merge all the way from l, l + 1, ... r - 1 -> r
            while (l != r) {
                merge(l, r);
                count--;
                l = find(l + 1);
            }
            res.push_back(count);
        }
        return res;
    }

    void merge(int l, int r) { nodes[l] = r; }

    int find(int node) {
        if (node != nodes[node]) {
            nodes[node] = find(nodes[node]);
        }
        return nodes[node];
    }
};