// 1376. Time Needed to Inform All Employees #include #include #include #include /* * a company has 'n' employees with a unique id for each employee from 0 to 'n - * 1'. the head of the company is the one with the 'head_id'. each employee has * one direct manager given the 'manager' array where 'manager[i]' is the direct * manager of the i'th employee, 'manager[head_id] = -1'. also it is guaranteed * that the subordination relations have a tree structure. the head of the * company wants to inform all the company employees of an urgent piece of new. * he will inform his direct subordinates, and they will inform their * subordinates and so on until all the employees know about the urgent news. * the i'th employee needs 'inform_time[i]' minutes to inform all of his direct * subordinates (ie. after 'inform_time[i]' minutes, all his direct subordinates * can start spreading the news) return the number of minutes needed to inform * all the employees about the urget news. */ int numOfMinutes(int n, int head_id, int *manager, int manager_size, int *inform_time, int inform_time_size) { if (n == 1) return 0; int *cn = calloc(n, sizeof(int)); for (int i = 0; i < n; i++) { if (i == head_id) continue; cn[manager[i]]++; } int *min_time = malloc(n * sizeof(int)); for (int i = 0; i < n; i++) min_time[i] = INT_MIN; int *stack = malloc(n * sizeof(int)); int head = 0, idx = 0; for (int i = 0; i < n; i++) if (!cn[i]) { min_time[i] = 0; stack[idx] = i; idx++; } while (head < idx) { int emp = stack[head], mag = manager[emp]; head++; if (mag >= 0) { cn[mag]--; min_time[mag] = fmax(min_time[mag], inform_time[mag] + min_time[emp]); if (!cn[mag]) { stack[idx] = mag; idx++; } } } int ans = min_time[head_id]; free(cn), free(stack), free(min_time); return ans; } int main() { int m1[] = {-1}, m2[] = {2, 2, -1, 2, 2, 2}; int it1[] = {0}, it2[] = {0, 0, 1, 0, 0, 0}; printf("%d\n", numOfMinutes(1, 0, m1, 1, it1, 1)); // expect: 0 printf("%d\n", numOfMinutes(6, 2, m2, 6, it2, 6)); // expect: 1 }