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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
-}