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// 2050. Parallel Courses III
#include "leetcode.h"
/*
* given an integer 'n' which indicates that there are n courses labeled from 1
* to n. you are also given a 2d integer array rleations where relations[j] =
* pre_course[j], denotes that course pre_course has to be completed before
* next_course. furthermore, you are given a 0-indexed integer array 'time'
* where 'time[i]' denotes how many months it takes to complete the '(i + 1)'th
* course. you must find the minimum number of months needed to complete all the
* courses following these rules. you may start taking courses at any time if
* the prerequisites have been met, and any number of courses can be taken at
* the same time. return the minimum number of monthsneeded to omplete all the
* courses.
*/
class Solution {
public:
int minimumTime(int n, vvd & relations, vector<int> &time) {
vector<int> in_deg(n + 1);
vvd g(n + 1);
for (auto r : relations) {
int a = r[0];
int b = r[1];
g[a].push_back(b);
in_deg[b]++;
}
vector<int> finish_time(n + 1);
queue<int> q;
for (int i = 1; i <= n; i++)
if (!in_deg[i])
q.push(i);
while (!q.empty()) {
int curr = q.front();
q.pop();
finish_time[curr] += time[curr - 1];
for (auto next : g[curr]) {
in_deg[next]--;
finish_time[next] = max(finish_time[next], finish_time[curr]);
if (!in_deg[next])
q.push(next);
}
}
return *max_element(finish_time.begin(), finish_time.end());
}
};
int main() {
Solution obj;
vvd r1 = {{1, 3}, {2, 3}}, r2 = {{1, 5}, {2, 5}, {3, 5}, {3, 4}, {4, 5}};
vector<int> t1 = {3, 2, 5}, t2 = {1, 2, 3, 4, 5};
printf("%d\n", obj.minimumTime(3, r1, t1)); // expect: 8
printf("%d\n", obj.minimumTime(5, r2, t2)); // expect: 12
}
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