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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(int) & relations, vector<int> &time) {
+ vector<int> in_deg(n + 1);
+ vvd(int) 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(int) 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
+}