解決思路:
利用遞歸,1...N個物品,每種有Amount[i]個,那麼從第N個物品開始選擇,選0個、...、Amount[i]個第N個物品,依次計算價值,找出最大的,傳回。
F(N) = max{ F(N-1, Lmt-n*Cost(n)) + n*Value(n), | n=0,..., Amount[i]}。
#ifndef HJD_SNACKBACK_H_
#define HJD_SNACKBACK_H_
int array_Weight[5] = {10, 5, 3, 6, 2};
int array_Vacancy[5] = {5, 15, 8, 6, 5};
int array_Amount[5] = {3, 3, 3, 5, 7};
int array_Price[5] = {6, 10, 8, 6, 15};
int Weight_Lmt = 30;
int Vacancy_Lmt =30;
//遞歸去計算
int hjd_complete_snackback(int nPos, int nWeightLmt, int nVacancyLmt)
{
int nAmount=0, nResult=0;
if(nPos==0)
{
for(nAmount=0;
(nAmount<=array_Amount[nPos])&&(nAmount*array_Weight[nPos]<=nWeightLmt)&&(nAmount*array_Vacancy[nPos]<=nVacancyLmt);
nAmount++)
{
nResult=nAmount*array_Price[nPos];
}
}
else
{
nAmount = 0;
int ntmpResult = 0;
while((nAmount<=array_Amount[nPos])&&(nAmount*array_Weight[nPos]<=nWeightLmt)&&(nAmount*array_Vacancy[nPos]<=nVacancyLmt))
{
ntmpResult = hjd_complete_snackback(nPos-1,
nWeightLmt - nAmount*array_Weight[nPos],
nVacancyLmt - nAmount*array_Weight[nPos]) + nAmount*array_Price[nPos];
if (nResult<ntmpResult)
nResult = ntmpResult;
nAmount++;
}
}
return(nResult);
}
#endif /* HJD_SNACKBACK_H_ */
測試正确。