2.1 holstein

好久没有做题了。

此题的关键是组合的产生，细想一下也是可以用递归做的。

我使用了简单的模拟。

 

Healthy Holsteins
Burch & Kolstad

Farmer John prides himself on having the healthiest dairy cowsin the world. He knows the vitamin content for one scoop of eachfeed type and the minimum daily vitamin requirement for the cows.Help Farmer John feed his cows so they stay healthy while minimizingthe number of scoops that a cow is fed.

Given the daily requirements of each kind of vitamin that a cowneeds, identify the smallest combination of scoops of feed a cowcan be fed in order to meet at least the minimum vitamin requirements.

Vitamins are measured in integer units. Cows can be fed at mostone scoop of any feed type. It is guaranteed that a solution existsfor all contest input data.

PROGRAM NAME: holstein

INPUT FORMAT

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.

SAMPLE INPUT (file holstein.in)

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

OUTPUT FORMAT

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

If more than one set of feedtypes yield a minimum of scoops, choosethe set with the smallest feedtype numbers.

SAMPLE OUTPUT (file holstein.out)

2 1 3

/*ID: xdzhbin1LANG: C++TASK: holstein*/#include <fstream>#include <iostream>#include <vector>using namespace std;//判断一个组合是否满足条件bool check(int nElem, int need[], int feed[][25], int vit[], int r, int c){int i,j;int sum = 0;for(i=0; i<c; i++){sum = 0;for (j=0; j<=nElem; j++){sum += feed[need[j]-1][i];}if (vit[i]>sum){return false;}}return true;}//产生组合的顺序//nElems，元素的个数//need存放组合//r全集中元素的个数void update(int& nElems, int need[], int r, int c){int i ;int bits = 0;need[nElems]++;while (need[nElems] > r)//需要进位{bits++;if (nElems - bits < 0) //组合位数增加{nElems++;for (i=0; i<=nElems; i++){need[i] = i+1;}break;}else{need[nElems-bits]++;for (i = nElems-bits+1; i<=nElems; i++){need[i] = need[i-1] + 1;}}}}int main(){int nVitamins;int nFeeds;int vit[25];int feed[15][25];int i,j;for (i=0; i<25; i++){vit[i] = 0;}for (i=0; i<15; i++){for (j=0; j<25; j++){feed[i][j] = 0;}}ifstream in("holstein.in");ofstream out("holstein.out");in>>nVitamins;for (i=0; i<nVitamins; i++){in>>vit[i];}in>>nFeeds;for (i=0; i<nFeeds; i++){for (j=0; j<nVitamins; j++){in>>feed[i][j];}}bool enough = false;int nElem = 0;int needs[15];for (i=0; i<15; i++){needs[i] = 0;}while (!enough){enough = check(nElem,needs,feed,vit, nFeeds, nVitamins);if (!enough){update(nElem,needs,nFeeds,nVitamins);}/*for (i=0;i<=nElem; i++){out<<needs[i]<<" ";}out<<endl;*/}out<<nElem+1<<" "<<needs[0];for (i=1; i<=nElem; i++){out<<" "<<needs[i];}out<<endl;return 0;}