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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 C implementation makes use of a recursive DFS function to calculate the minimum completion time for each course by traversing through its dependencies. Memorization is utilized to store the time taken to complete each course dp[u], thus optimizing the solution by preventing recalculations.