USACO 3.3 Shopping Offers

USACO 3.3 Shopping Offers

背包问题，dp解决。
dp[i][j][k][m][n]表示第0-4种物品分别购买i-n个时，所需的最小费用。
那么对于每一个offer
dp[i][j][k][m][n] = min ( dp[i][j][k][m][n], dp[i-o[0]][j-o[1]][k-o[2]][m-o[3]][n-o[4]]+offer.cost);

analysis中提出用最短路径的方法来解，思路很巧妙。把”装有不同种数和数量的物品“的篮子看作结点，把offer作为边(把购买一件物品的原始价格看成一种退化的offer)，把价格作为边长度。这样就转换成从篮子(0,0,0,0,0)到所求结点的最短路径问题。

代码如下:

#include  < iostream >
#include  < fstream >

using   namespace  std;

ifstream fin( " shopping.in " );
ofstream fout( " shopping.out " );

#ifdef _DEBUG
#define  out cout
#define  in cin
#else
#define  out fout
#define  in fin
#endif

int  dp[ 6 ][ 6 ][ 6 ][ 6 ][ 6 ];

struct  offer{
     int  pro_num;
     int  price;
     int  product[ 5 ];
     int  amount[ 5 ];
};

int  offer_num;
offer offers[ 100 ];
int  map[ 1000 ]; //product id到”按物品出现顺序所给的编号“的映射
int  price[ 1000 ];//product id对应的物品价格
int  product_num;//物品总数目
int  products[ 5 ];//存放product id
int  amount[ 5 ];//product 所需的数量

void  solve()
{
     in >> offer_num;

     for ( int  i = 0 ;i < offer_num; ++ i){
         in >> offers[i].pro_num;
         for ( int  j = 0 ;j < offers[i].pro_num; ++ j){
             in >> offers[i].product[j] >> offers[i].amount[j];
        }
         in >> offers[i].price;
    }
    
     int  pro_idx  =   0 ;

     in >> product_num;

     for ( int  i = 0 ;i < product_num; ++ i){
         in >> products[i];
         in >> amount[i];
         in >> price[i];
        map[ products[i] ]  =  i;
    }

    //没有折扣时的价格
     for ( int  i = 0 ;i <= 5 ; ++ i)
         for ( int  j = 0 ;j <= 5 ; ++ j)
             for ( int  k = 0 ;k <= 5 ; ++ k)
                 for ( int  m = 0 ;m <= 5 ; ++ m)
                     for ( int  n = 0 ;n <= 5 ; ++ n){
                        dp[i][j][k][m][n]  =  
                            price[ 0 ] * i +
                            price[ 1 ] * j +
                            price[ 2 ] * k +
                            price[ 3 ] * m +
                            price[ 4 ] * n;
                    }

     for ( int  i = 0 ;i <= 5 ; ++ i)
         for ( int  j = 0 ;j <= 5 ; ++ j)
             for ( int  k = 0 ;k <= 5 ; ++ k)
                 for ( int  m = 0 ;m <= 5 ; ++ m)
                     for ( int  n = 0 ;n <= 5 ; ++ n){
                         int  tmp[ 5 ];
                         for ( int  s = 0 ;s < offer_num; ++ s){

                            memset(tmp, 0 , sizeof (tmp));    
                             for ( int  t = 0 ;t < offers[s].pro_num; ++ t){
                                tmp[map[offers[s].product[t]]]  =  offers[s].amount[t];
                            }

                             if (i >= tmp[ 0 ] && j >= tmp[ 1 ] && k >= tmp[ 2 ] && m >= tmp[ 3 ] && n >= tmp[ 4 ]){
                                dp[i][j][k][m][n]  =  min( dp[i][j][k][m][n],
                                        dp[i - tmp[ 0 ]][j - tmp[ 1 ]][k - tmp[ 2 ]][m - tmp[ 3 ]][n - tmp[ 4 ]] +
                                        offers[s].price);
                            }
                        }
                    }

     out << dp[amount[ 0 ]][amount[ 1 ]][amount[ 2 ]][amount[ 3 ]][amount[ 4 ]] << endl;
}

int  main( int  argc, char   * argv[])
{
    solve(); 
     return   0 ;
}

上面代码有个严重的bug，谢谢网友“我也凑热闹"指出。由于map所有值都为0。所以未在商品列表中出现的商品的map值都为0，即都映射为第一个商品。现改成将map初始化为-1，并增加判断语句。此外将初始化dp的语句合并到后面，以简化代码。

#include  < iostream >
#include  < fstream >

using   namespace  std;

ifstream fin( " shopping.in " );
ofstream fout( " shopping.out " );

#ifdef _DEBUG
#define  out cout
#define  in cin
#else
#define  out fout
#define  in fin
#endif

int  dp[ 6 ][ 6 ][ 6 ][ 6 ][ 6 ];

struct  offer{
     int  pro_num;
     int  price;
     int  product[ 5 ];
     int  amount[ 5 ];
};

int  offer_num;
offer offers[ 100 ];
int  map[ 1000 ];
int  price[ 1000 ];
int  product_num;
int  products[ 5 ];
int  amount[ 5 ];

void  solve()
{
     in >> offer_num;

     for ( int  i = 0 ;i < offer_num; ++ i){
         in >> offers[i].pro_num;
         for ( int  j = 0 ;j < offers[i].pro_num; ++ j){
             in >> offers[i].product[j] >> offers[i].amount[j];
        }
         in >> offers[i].price;
    }
    
     int  pro_idx  =   0 ;

     in >> product_num;

     // 2009.07.27修改
    memset(map, - 1 , sizeof (map));

     for ( int  i = 0 ;i < product_num; ++ i){
         in >> products[i];
         in >> amount[i];
         in >> price[i];
        map[ products[i] ]  =  i;
    }

     for ( int  i = 0 ;i <= 5 ; ++ i)
         for ( int  j = 0 ;j <= 5 ; ++ j)
             for ( int  k = 0 ;k <= 5 ; ++ k)
                 for ( int  m = 0 ;m <= 5 ; ++ m)
                     for ( int  n = 0 ;n <= 5 ; ++ n){

                         dp[i][j][k][m][n]  =  
                            price[ 0 ] * i +
                            price[ 1 ] * j +
                            price[ 2 ] * k +
                            price[ 3 ] * m +
                            price[ 4 ] * n;

                         int  tmp[ 5 ];
                         for ( int  s = 0 ;s < offer_num; ++ s){
                            memset(tmp, 0 , sizeof (tmp));    
                             for ( int  t = 0 ;t < offers[s].pro_num; ++ t){
                                 // 2009.07.27修改
                                 if ( map[offers[s].product[t]] !=- 1 )
                                    tmp[map[offers[s].product[t]]]  =  offers[s].amount[t];
                            }

                             if (i >= tmp[ 0 ] && j >= tmp[ 1 ] && k >= tmp[ 2 ] && m >= tmp[ 3 ] && n >= tmp[ 4 ]){
                                dp[i][j][k][m][n]  =  min( dp[i][j][k][m][n],
                                        dp[i - tmp[ 0 ]][j - tmp[ 1 ]][k - tmp[ 2 ]][m - tmp[ 3 ]][n - tmp[ 4 ]] +
                                        offers[s].price);
                            }
                        }
                    }

     out << dp[amount[ 0 ]][amount[ 1 ]][amount[ 2 ]][amount[ 3 ]][amount[ 4 ]] << endl;
}

int  main( int  argc, char   * argv[])
{
    solve(); 
     return   0 ;
}