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This approach involves topologically sorting the courses based on their prerequisites and then calculating the minimal time to complete each course using dynamic programming.
(a,b) is a directed edge from a to b.Time Complexity: O(n + E), where n is the number of courses and E is the number of prerequisite relationships, due to traversing all nodes and edges once.
Space Complexity: O(n + E), needed to store the graph and additional arrays.
1```cpp
2#include <iostream>
3#include <vector>
4#include <queue>
5#include <algorithm>
6
7using namespace std;
8
9int minTime(int n, vector<vector<int>>& relations, vector<int>& time) {
10 vector<int> adj[n+1];
11 vector<int> indegree(n+1, 0);
12 vector<int> dp(n+1, 0);
13 for (const auto& rel : relations) {
14 adj[rel[0]].push_back(rel[1]);
indegree[rel[1]]++;
}
queue<int> q;
for (int i = 1; i <= n; ++i) {
if (indegree[i] == 0) {
q.push(i);
dp[i] = time[i-1];
}
}
while (!q.empty()) {
int u = q.front(); q.pop();
for (int v : adj[u]) {
dp[v] = max(dp[v], dp[u] + time[v-1]);
if (--indegree[v] == 0) {
q.push(v);
}
}
}
return *max_element(dp.begin(), dp.end());
}
int main() {
int n = 5;
vector<vector<int>> relations = {{1,5},{2,5},{3,5},{3,4},{4,5}};
vector<int> time = {1, 2, 3, 4, 5};
cout << minTime(n, relations, time) << endl;
return 0;
}
```
The C++ solution leverages vectors to represent the adjacency lists and includes the indegree tracking for each course. It performs a topological sort by using a queue (BFS manner) to process courses whose prerequisites are fulfilled. The dp array is dynamically updated. Finally, it determines the maximum time required to complete all courses.
This approach uses Depth-First Search (DFS) with memorization to efficiently compute the completion time for courses.
Time Complexity: O(n + E), visiting each node and edge at least once.
Space Complexity: O(n + E), due to storage for graph data and function call stack.
```This Python implementation effectively uses DFS with a dictionary adj to trace course dependencies, and a dp array to memorize computed outcomes. The result is determined by calculating the maximum time taken from these evaluated paths.