USACO 3.3.2 Shopping Offers (5维完全背包)

USACO 3.3.2 Shopping Offers (5维完全背包)

最多有5种商品，把每一种优惠，把他们看成分别有5个权重（w1,w2,w3,w4,w5），一种价值cost的背包。

另外也把原价购买一件商品作为一种背包，例如（w1=1，w2..5=0）cost=3
 

/**/ /*
ID: lorelei3
TASK: shopping
LANG: C++
*/

#include  < fstream >

using   namespace  std;

const   int  MAX  =   100 ;
const   int  INF  =   999999999 ;
typedef  struct {
    int w[6];
    int cost;
} Offer;

Offer offers[ 100 ];

int  a[ 6 ], hash[ 10 ];
int  s,m,t,c,k,p,n,b;
int  v1,v2,v3,v4,v5;

int  f[ 6 ][ 6 ][ 6 ][ 6 ][ 6 ];

inline  int  decode( int  code) {
    int i;
    for(i=1; i<=t; ++i)
        if(hash[i]==code)
            return i;
    hash[++t] = code;
    return t;
}

int  main() {
    int i,j;

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

    fin>>s;

    for(i=0; i<s; ++i){
        ++m;
        fin>>n;
        for(j=0; j<n; ++j){
            fin>>c>>k;
            offers[m].w[decode(c)] = k;
        }
        fin>>offers[m].cost;
    }

    fin>>b;

    for(v1=0; v1<6; ++v1)
        for(v2=0; v2<6; ++v2)
            for(v3=0; v3<6; ++v3)
                for(v4=0; v4<6; ++v4)
                    for(v5=0; v5<6; ++v5)
                        f[v1][v2][v3][v4][v5]=INF;
    f[0][0][0][0][0] = 0 ;

    for(i=0; i<b; ++i){
        fin>>c>>k>>p;
        ++m;
        int code = decode(c);
        offers[m].w[code] = 1;
        offers[m].cost = p;
        a[code] = k;
    }

    for(i=1; i<=m; ++i)
        for(v1=offers[i].w[1]; v1<=5; ++v1)
            for(v2=offers[i].w[2]; v2<=5; ++v2)
                for(v3=offers[i].w[3]; v3<=5; ++v3)
                    for(v4=offers[i].w[4]; v4<=5; ++v4)
                        for(v5=offers[i].w[5]; v5<=5; ++v5)
                            if(f[v1][v2][v3][v4][v5] > f[v1-offers[i].w[1]][v2-offers[i].w[2]][v3-offers[i].w[3]][v4-offers[i].w[4]][v5-offers[i].w[5]] + offers[i].cost )
                                f[v1][v2][v3][v4][v5] = f[v1-offers[i].w[1]][v2-offers[i].w[2]][v3-offers[i].w[3]][v4-offers[i].w[4]][v5-offers[i].w[5]] + offers[i].cost;
                        

    fout<<f[a[1]][a[2]][a[3]][a[4]][a[5]]<<endl;

    return 0;
}

PS:最近忙着写标书，真是廉价劳动力哇。。