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// 1376. Time Needed to Inform All Employees
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
/*
* 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
}
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