3243. Shortest Distance After Road Addition Queries I

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3243. Shortest Distance After Road Addition Queries I

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.

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 <= 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.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    vector<vector<int>> graph;
    vector<bool> visited;
    queue<int> q;

    vector<int> shortestDistanceAfterQueries(int n,
                                             vector<vector<int>>& queries) {
        graph.resize(n, vector<int>());
        for (int i = 0; i < n; i++) {
            if (i < n - 1) {
                graph[i].push_back(i + 1);
            }
        }
        vector<int> res;
        for (auto& query : queries) {
            int u = query[0];
            int v = query[1];
            graph[u].push_back(v);
            visited = vector<bool>(n, false);
            q = queue<int>();
            cut(0);
            int cur = 0;
            bool finished = false;
            while (!q.empty()) {
                cur++;
                int size = q.size();
                for (int i = 0; i < size; i++) {
                    int u = q.front();
                    q.pop();
                    if (u == n - 1) {
                        res.push_back(cur);
                        finished = true;
                        break;
                    }
                    cut(u);
                }
                if (finished) {
                    break;
                }
            }
        }
        return res;
    }

    void cut(int u) {
        for (int v : graph[u]) {
            if (!visited[v]) {
                visited[v] = true;
                q.push(v);
            }
        }
    }
};