USACO3.3.3DP

/*ID: 19930323PROG: shoppingLANG: C++HINT：官方题解是最短路，我的思路是DP；先应将商品编号离散至0~b-1；商品数只有5种，最多每种商品买5个，因此用到六进制存储打包促销方案；如:官方样例中打包方案"1 7 3 5"存储为item=3,price=5;方案"2 7 1 8 2 10"存储为item=1+6*2=11,price=10;另外原价销售如"7 3 2"存储为item=1,price=2,"8 2 5"存储为item=6*1,price=5;记f[x1*1+x2*6+x3*6^3+x4*6^4+x5*6^5]为买想x1件商品1,x2件商品2,x3件商品3,x4件商品4,x5件商品5所需最少的钱数；记goal=k1*1+k2*6+k3*6^3+k4*6^4+k5*6^5,ki为要买的商品i的数量；接下来就是从f[0]到f[goal]的DP过程了，有状态转移方程f[i]=min{f[i-p[j].item]+p[j].price}(i-p[j].item>=0)，这个过程见代码；DP过程可换做图论求最短路过程，实质是一样的。*/#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int INF=0x7fffffff;const int ep[5]={1,6,36,216,1296};struct node{    int item;    int price;}p[120];int id[1000];int f[8000];  //事实上状态数最多为0~7775int t,idcnt,goal;int get_id(int x){    if (id[x]!=-1) return id[x];    else return id[x]=idcnt++;}void init(){    idcnt=0;    memset(id,-1,sizeof(id));    int s;    scanf("%d",&s);    t=0;    for (int i=1; i<=s; i++,t++)    {        p[t].item=0;        int k;        scanf("%d",&k);        for (int j=1; j<=k; j++)        {            int x,y,z;            scanf("%d%d",&x,&y);            z=get_id(x);            p[t].item+=ep[z]*y;        }        scanf("%d",&p[t].price);    }    int b;    scanf("%d",&b);    goal=0;    for (int i=1; i<=b; i++,t++)    {        int c,k,pp;        scanf("%d%d%d",&c,&k,&pp);        int x=get_id(c);        p[t].item=ep[x];        p[t].price=pp;        goal+=k*ep[x];    }}void solve(){    for (int i=1; i<=goal; i++) f[i]=INF;    f[0]=0;    for (int i=0; i<goal; i++)    {        for (int j=0; j<t; j++)        {            int k=i+p[j].item;            if (k>goal) continue;            f[k]=min(f[k],f[i]+p[j].price);        }    }    printf("%d\n",f[goal]);}int main(){    freopen("shopping.in","r",stdin);    freopen("shopping.out","w",stdout);    init();    solve();    return 0;}