[USACO3.3.2 Shopping Offers]

[关键字]：动态规划

[题目大意]：你要买n种商品，每种买k件。同时有一些组合购买的优惠方案，问最小花费。

//====================================================================================================

[分析]：因为最多才有五种商品，每种最多5件，所以五维的动态规划完全可以接受。至于方程太麻烦了，不写了，代码里很清楚……

[代码]：

View Code

/*
ID:procedure2
PROB:shopping
LANG:C++
*/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int INF=1684300900;

struct node
{
       int sum,pay;
       int next[10];
}a[100],d[100];
int s,n,m;
int b[1000];
int f[11][11][11][11][11];

void Init()
{
    scanf("%d",&s);
    for (int i=1;i<=s;i++)
    {
        scanf("%d",&a[i].sum);
        //printf("%d\n",a[i].sum);
        for (int j=1;j<=a[i].sum;j++)
        {
            int c,k;
            scanf("%d%d",&c,&k);
            if (!b[c]) b[c]=++n;
            a[i].next[b[c]]=k;
        }
        scanf("%d",&a[i].pay);
        //printf("%d %d %d %d %d %d %d\n",a[i].sum,a[i].next[1],a[i].next[2],a[i].next[3],a[i].next[4],a[i].next[5],a[i].pay);
    }
    //for (int i=1;i<=s;i++) printf("%d %d %d %d %d %d %d\n",a[i].sum,a[i].next[1],a[i].next[2],a[i].next[3],a[i].next[4],a[i].next[5],a[i].pay);
    scanf("%d",&m);
    for (int i=1;i<=m;i++)
    {
        int c,k;
        scanf("%d%d",&c,&k);
        if (!b[c]) b[c]=++n;
        d[b[c]].sum=k;
        scanf("%d",&d[b[c]].pay);
    }
    for (int i=1;i<=n;i++)
    {
        a[s+i].sum=1;
        a[s+i].next[i]=1;
        a[s+i].pay=d[i].pay;
    }
    s+=n;
    //for (int i=1;i<=s;i++) printf("%d %d %d %d %d %d %d\n",a[i].sum,a[i].next[1],a[i].next[2],a[i].next[3],a[i].next[4],a[i].next[5],a[i].pay);
}

void Solve()
{
     memset(f,100,sizeof(f));
     f[0][0][0][0][0]=0;
     for (int i1=0;i1<=d[1].sum;i1++)
         for (int i2=0;i2<=d[2].sum;i2++)
             for (int i3=0;i3<=d[3].sum;i3++)
                 for (int i4=0;i4<=d[4].sum;i4++)
                     for (int i5=0;i5<=d[5].sum;i5++)
                         if (f[i1][i2][i3][i4][i5]!=INF)
                            for (int j=1;j<=s;j++)
                                f[i1+a[j].next[1]][i2+a[j].next[2]][i3+a[j].next[3]][i4+a[j].next[4]][i5+a[j].next[5]]
                                =min(f[i1+a[j].next[1]][i2+a[j].next[2]][i3+a[j].next[3]][i4+a[j].next[4]][i5+a[j].next[5]],f[i1][i2][i3][i4][i5]+a[j].pay);
     printf("%d\n",f[d[1].sum][d[2].sum][d[3].sum][d[4].sum][d[5].sum]);
}

int main()
{
    freopen("shopping.in","r",stdin);
    freopen("shopping.out","w",stdout);
    Init();
    Solve();
    return 0;
}