REIGN2 - Editorial

PROBLEM LINK:

Practice
Contest

Author: Kanstantsin Sokal
Tester: Jingbo Shang
Editorialist: Lalit Kundu

DIFFICULTY:

Easy-Medium

PREREQUISITES:

dynamic programming, greedy

PROBLEM:

There are N castles that can be conquered by George. He has decided to conquer exactly K castles during the next K days, an aim he plans to achieve by conquering one castle every day for the next K days.
As reported by King George’s spies, A_i + (j - 1) * B_i enemy knights will be protecting the i^{th} castle during the j^{th} day. When attacking a castle, if the King sends as many knights as those defending it, it’s sufficient to be conquer that castle. Another requirement is that one knight cannot be sent to conquer two different castles.

As you are the king’s trusted advisor, you can choose a set of K castles, that the king should conquer. After you provide a set of castles, George will choose the order in which he will conquer them. You can assume that George always chooses the order which minimizes the total number of knights used during all the conquests.

Your task is to find the maximum possible total number of knights that can be used during the campaign and reached by delegating a set of K castles to conquer. Also, you are not sure about the value of K, so you should find the optimal answer for each K = 1, 2, ... , N.

QUICK EXPLANATION:

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First sort arrays A and B in decreasing order of B while preserving the mutual connectedness of A_i and B_i. After that, we define f(i, j) as maximum number of knights required for king to conquer a subset of j castles from first i castles. We can easily define recurrence relation and base cases of this DP solution.

EXPLANATION:

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First of all let’s observe the strategy of king. He is given a set of K castles and two arrays A_1, A_2, ..., A_K and B_1, B_2, ..., B_K are given. For conquering castle i on day j he has to send A_i + (j - 1)*B_i knights. His aim is to decide such an ordering of conquering such that total number of knights required is minimised.

Now, he has to pay A_i for each castle i independent of which day he conqueres it on. So, the ordering depends only on B_i. If we think of a greedy approach we should first conquer those castles with highest B_i since, as the day number increases (j-1)*B_i will also increase. So, this is the the strategy of king: Conquer all K castles in non-increasing order of B_i.

Now, moving onto solving the problem. Let’s solve for a particular K. So, we have to choose K castles out of N such that using king’s approach maximum number of knights are used. Now, a greedy approach which selects only of basis of A_i or B_i would be certainly wrong. We should, by now, be thinking on terms of dynamic programming.

Now, if you remember subset sum DP, we kept our states as (i, j) denoting that we are solving for first i elements and we need to form a sum j. If we try to think on similar terms here, we should keep our states as (i, j) denoting that we are solving for first i elements and we need to select j castles to maximise knights required according to king’s strategy. Now, we have to make a decision that either we select i^{th} castle or not. If we select it, do we know what its going to contribute i.e. how many knights are going to be required? For that we should know on which day king will conquer this castle. Now, here is the interesting part: if we sort the initial arrays in decreasing order of B_i, then we’ll be sure that according to king’s strategy i^{th} castle will be conquered on last day because all other castles that will be selected will have B_j less than B_i.

So, we first sort arrays A and B in decreasing order of B while preserving the mutual connectedness of A_i and B_i. After that, we define f(i, j) as maximum number of knights required for king to conquer a subset of j castles from first i castles.

Recurrences can be define easily as:

f(i, j) = max(
		\\don't select i'th castle
		f(i - 1, j)	

		\\select i'th castle, add the cost
		f(i - 1, j - 1) + (A* + (j - 1)*B*)		
	)

Realising base cases is very easy. So, we calculate f(N-1, i) for all i = 1, 2, ..., N and print them.

COMPLEXITY:

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Since there are N^2 states and transition cost between states is constant time, total complexity is O(N^2).

PROBLEMS TO SOLVE:

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KGP14D
KGP14H
MCHEF
CARDLINE

AUTHOR’S, TESTER’S SOLUTIONS:

setter
tester

5 Likes

Can you please explain the first sample case ?

Cant view the tester and setters solution, please fix it.

For the second case :

for k = {1, 2} the knights in castles are :

K = 1 : 4, 6, 4, 4 3

K = 2 : 13, 12, 11, 13, 11

So for K = 2, how can answer be 17. I think the answer is 19 because in day 1, select 2nd castle and in day 2, select 1st castle .So 6 + 13 = 19).

Can any one explain this.

Links for the solution not working. Please check.

1 Like

For K = 1
Its optimal to choose the 4th castle as the king has only 1 option to send 5 knights. Any other castle will require less knights
For K = 2
If we choose the 1st and 5th knight , the king will send 5 + 3 + 1 * 4 = 12 knights , any other pair of castle will result in less knights being sent
For K = 3
If we choose the 1st , 2nd and 4th knight , the king will spend 5 * 0 + 1 + 4 * 1 + 3 + 5 + 4 * 2 = 21 knights
For K = 4
We have no choice but to choose all the castles , so the king will spend (1 + 5 * 0) + (3 + 4 * 1) + (5 + 4 * 2) + (1 + 3 * 3) = 31 knights