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-// 3243. Shortest Distance After Road Addition Queries I
-#include "leetcode.h"
-
-/*
- * 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] = [u_i,
- * v_i]' represents the addition of a new unidirectional road from city 'u_i' to
- * city 'v_i'. 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 of '[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.
- */
-
-class Solution {
-public:
- vector<int> shortestDistanceAfterQueries(int n,
- vector<vector<int>> &queries) {
- vector<int> ans;
- unordered_map<int, vector<int>> mp;
- for (int i = 0; i < n - 1; i++)
- mp[i].push_back(i + 1);
- for (auto i : queries) {
- mp[i[0]].push_back(i[1]);
- vector<int> dp(n, INT_MAX);
- dp[n - 1] = 0;
- for (int j = n - 2; j >= 0; --j)
- for (int next : mp[j])
- dp[j] = min(dp[j], dp[next] + 1);
- ans.push_back(dp[0]);
- }
- return ans;
- }
-};
-
-int main() {
- Solution obj;
- vvd(int) q1 = {{2, 4}, {0, 2}, {0, 4}}, q2 = {{0, 3}, {0, 3}};
- vector<int> sdaq1 = obj.shortestDistanceAfterQueries(5, q1);
- vector<int> sdaq2 = obj.shortestDistanceAfterQueries(4, q2);
- for (auto i : sdaq1)
- printf("%d ", i); // expect: 3 2 1
- printf("\n");
- for (auto i : sdaq2)
- printf("%d ", i); // expect: 1 1
- printf("\n");
-}