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IPCTRAIN - Editorial

Primary Tester: Misha Chorniy
Editorialist: Hussain Kara Fallah

Easy

Greedy,Heap

PROBLEM:

You have an upcoming camp. There are N trainers. The camp runs for D days.Each day,there can be at most one lecture. The ith trainer arrives on day Di and then stays till the end of the camp. He also wants to teach exactly Ti lectures. For each lecture that a trainer was not able to teach,his sadness level will be increased by Si. You are the main organizer of the contest. You want to find minimum total sadness of the trainers.

EXPLANATION:

Let's assign our trainers starting from the first day of the camp. At each day adding arriving trainers to our set of trainers.

Each day should be assigned to only one trainer, it's obvious that we should assign the trainer with maximum sadness to this lecture, so our set (container) should be a heap sorting trainers by their sadness value. In fact we should know also the number of days each lecturer wants to serve in, so we take the lecturer with maximum sadness from our heap and decrease his demand by 1. In case he still has lectures he would like to do, we keep him in the heap, otherwise we just pop him out.

After finishing all days, some teachers may have some remaining lectures they wanted to teach but we couldn't find such days for them. Let's denote the number of lectures the ith trainer couldn't teach by remi So our answer would be :

$answer = \sum_{i=1}^{i=N} rem_i * S_i$

AUTHOR'S AND TESTER'S SOLUTIONS:

AUTHOR's solution: Will be found here
TESTER's solution: Will be found here
EDITORIALIST's solution: Will be found here

This question is marked "community wiki".

118823
accept rate: 0%

 6 This question literally screamed "PRIORITY QUEUE" in my ears. One of the good questions which can be solved with knowledge of proper data type!! Honestly, i saw priority queue code for first time in "Cooking Schedule" editorial (one of the answers told that "it can also be solved using priority queue") and it made life so simple for me here. answered 17 Jul '17, 15:30 14.1k●1●13●48 accept rate: 18% 2 Solution : Great! Comments : Extraordinary :D (17 Jul '17, 15:38) Lol XD. It just came into my mind when i read "A trainer who comes to the camp stays there till end of camp" :p (17 Jul '17, 15:40)
 1 I have used Segment Trees answered 17 Jul '17, 16:52 11●1 accept rate: 0%
 1 @lord_ozb: You have used a double loop here which caused the time to exceed the limit.  for(int i = 0; i < d; i++){ int x = -1; int max = Integer.MIN_VALUE; for(int j = 0; j <= i; j++){ if(queue[j].size() != 0){ Trainer temp = queue[j].peek(); if(temp.sad > max){ max = temp.sad; x = j;  answered 18 Jul '17, 09:05 21●1 accept rate: 0%
 1 Can anyone help me? Author solution I seem that complexity O(N*T).Here N=100000 and T=100000 according to Constraints.How it's passed all test case.May be something i miss.Thanks in advance. answered 20 Jul '17, 20:47 2★aseem_cu 11●1 accept rate: 0%
 0 solutions are not available yet!! answered 17 Jul '17, 15:20 21●5 accept rate: 0%
 0 community - need help to understand what is missing https://www.codechef.com/viewsolution/14559868. only 40 pts answered 17 Jul '17, 18:58 21●1 accept rate: 0%
 0 Can anyone explain where I went wrong? 40 pts. https://www.codechef.com/viewsolution/14561290 answered 17 Jul '17, 21:26 3★lord_ozb 31●3 accept rate: 0%
 0 Hi! The link to practice and main contest are incorrect. Please update. answered 18 Jul '17, 08:32 1 accept rate: 0%
 0 can anyone explain why i m getting WA for last 2 subtask. https://www.codechef.com/viewsolution/14601668 answered 18 Jul '17, 09:05 1 accept rate: 0%
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question asked: 15 Jul '17, 07:23

question was seen: 4,757 times

last updated: 13 Jul, 10:10