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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 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");
}
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