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:

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 <= 105
• 1 <= queries.length <= 105
• 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].

Code

Time Complexity: $O(q\log n+n\log n)$, Space Complexity: $O(n)$

因為 There are no two queries such that i != j and queries[i][0] < queries[j][0] < queries[i][1] < queries[j][1] 所以可以用 set。

相反的，在 Shortest Distance After Road Addition Queries I 中因為不具備這個 constraint 因此必須使用 shortest path。

each element will only get insert and delete in the set at most twice.

class Solution {
public:
    vector<int> shortestDistanceAfterQueries(int n, vector<vector<int>>& queries) {
        set<int> st;
        for(int i = 0; i < n; i++) {
            st.insert(i);
        }
 
        vector<int> res;
        for(auto& q: queries) {
            int u = q[0];
            int v = q[1];
            remove_node_in_the_middle(st, u + 1, v - 1);
            res.push_back(st.size() - 1);
        }
 
        return res;
    }
 
    void remove_node_in_the_middle(set<int>& st, int u, int v) {
        auto l = st.lower_bound(u);
        auto r = st.upper_bound(v);
        st.erase(l, r);
    }
};

Source

• Shortest Distance After Road Addition Queries II - LeetCode