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-rw-r--r-- | shortest_dist_road_addition1.cpp | 48 | ||||
-rw-r--r-- | shortest_dist_road_addition1.py | 39 |
2 files changed, 87 insertions, 0 deletions
diff --git a/shortest_dist_road_addition1.cpp b/shortest_dist_road_addition1.cpp new file mode 100644 index 0000000..e0cc205 --- /dev/null +++ b/shortest_dist_road_addition1.cpp @@ -0,0 +1,48 @@ +// 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"); +} diff --git a/shortest_dist_road_addition1.py b/shortest_dist_road_addition1.py new file mode 100644 index 0000000..0115c4a --- /dev/null +++ b/shortest_dist_road_addition1.py @@ -0,0 +1,39 @@ +# 3243. Shortest Distance After Road Addition Queries I + +""" +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(object): + def shortestDistanceAfterQueries(self, n, queries): + """ + :type n: int + :type queries: List[List[int]] + :rtype: List[int] + """ + ans = [] + mp = {i: [i + 1] for i in range(n - 1)} + for i in queries: + mp[i[0]].append(i[1]) + dp = [float("inf")] * n + dp[n - 1] = 0 + for j in range(n - 2, -1, -1): + for next in mp[j]: + dp[j] = min(dp[j], dp[next] + 1) + ans.append(dp[0]) + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.shortestDistanceAfterQueries(n=5, queries=[[2, 4], [0, 2], [0, 4]])) + print(obj.shortestDistanceAfterQueries(n=4, queries=[[0, 3], [0, 2]])) |