Problem 2.1.6 Healthy Holsteins

출처: https://train.usaco.org/usacogate

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문제 설명

Farmer John prides himself on having the healthiest dairy cows in the world. He knows the vitamin content for one scoop of each feed type and the minimum daily vitamin requirement for his cows. Help Farmer John feed the cows so they stay healthy while minimizing the number of scoops that a cow is fed.
Given the daily requirements of each kind of vitamin that a cow needs, identify the smallest combination of scoops of feed a cow can be fed in order to meet at least the minimum vitamin requirements.
Vitamins are measured in integer units. Cows can be fed at most one scoop of any feed type. It is guaranteed that a solution exists for all contest input data.

입력

Line 1: integer V (1 <= V <= 25), the number of types of vitamins

Line 2: V integers (1 <= each one <= 1000), the minimum requirement for each of the V vitamins that a cow requires each day

Line 3: integer G (1 <= G <= 15), the number of types of feeds available

Lines 4..G+3:
V integers (0 <= each one <= 1000), the amount of each vitamin that one scoop of this feed contains. The first line of these G lines describes feed #1; the second line describes feed #2; and so on.

출력

The output is a single line of output that contains:
 - the minimum number of scoops a cow must eat, followed by:
 - a SORTED list (from smallest to largest) of the feed types the cow is given

입력 예

4
100 200 300 400
3
50 50 50 50
200 300 200 300
900 150 389 399

출력 예

2 1 3

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문제 풀이

모든 feed를 한번씩만 줄 수 있으므로, 전체 feed에서 가능한 모든 조합을 찾아서, 최소화 되는 조합을 출력한다.  feed 의 종류가 많지 않으므로( <= 15), bitset을 이용해서 필요한 조합을 찾는다. bitset<16> (0), ... bitset<16> (1<<numG) 다만, bitset으로 설정된 조합에서 작은 수들이 먼저 나오는 조합은 숫자로 변경하면, 더 큰 수가 된다.

프로그램 내용

...    
    for(int idx=0; idx < (1<<numG) ; ++idx)
    {
        vector <int> feeds(numV, 0);
        bitset<16> check(idx);

        for (int idx2=0; idx2 < numG ; ++idx2)
        {
            if( check[idx2] )
            {
                for(int idx3=0; idx3 < numV; ++idx3)
                {
                    feeds[idx3] += Feed_list[idx2][idx3];
                }
            }
        }

        /// check health : feeds vs V_req
        bool healthy = true;
        for(int idx2=0; idx2 < numV; ++idx2)
        {
            if ( feeds[idx2] < V_req[idx2])
            {
                healthy = false;
                break;
            }
        }

        if ( healthy 
        			&& check.count() < feed_count 
        			&& check.to_ulong() > feed_bit.to_ulong())
        {
            feed_count = check.count();
            feed_bit = check;
        }
    }
...

 

Chapter 2. Bigger Challenges (link)