diff options
Diffstat (limited to 'shortest_dist_road_addition1.cpp')
-rw-r--r-- | shortest_dist_road_addition1.cpp | 48 |
1 files changed, 0 insertions, 48 deletions
diff --git a/shortest_dist_road_addition1.cpp b/shortest_dist_road_addition1.cpp deleted file mode 100644 index e0cc205..0000000 --- a/shortest_dist_road_addition1.cpp +++ /dev/null @@ -1,48 +0,0 @@ -// 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"); -} |