USACO3.32Shopping Offers(DP)

五维DP，听着挺多的，貌似就是挺裸的dp，

最近貌似做简单的DP挺顺手。。1A

dp[i][j][e][o][g] = min(dp[i][j][e][o][g],dp[i-i1][j-i2][e-i3][o-i4][g-i5]+p[q])  i1,i2...为满足给出的商品数量的值 p[q]为选用当前优惠方案的价格。

 /*
     ID: shangca2
     LANG: C++
     TASK: shopping
  */
 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 using namespace std;
 #define INF 0xfffffff
 int dp[][][][][];
 struct node
 {
     int c[],k[],p,n;
 }pp[];
 int c[],k[],p[];
 int main()
 {
     freopen("shopping.in","r",stdin);
     freopen("shopping.out","w",stdout);
     int i,j,s,b,e,o,g,q,a;
     for(i = ; i <=  ; i++)
         for(j =  ; j <=  ; j++)
             for(e =  ; e <=  ; e++)
                 for(o =  ; o <=  ; o++)
                     for(g =  ; g <=  ; g++)
                     dp[i][j][e][o][g] = INF;
     cin>>s;
     for(i = ; i <= s ; i++)
     {
         cin>>pp[i].n;
         for(j = ; j <= pp[i].n ; j++)
         cin>>pp[i].c[j]>>pp[i].k[j];
         cin>>pp[i].p;
     }
     cin>>b;
     for(i =  ; i <= b ;i++)
     cin>>c[i]>>k[i]>>p[i];
     for(i =  ;i <= k[] ; i++)
     for(j = ; j <= k[] ; j++)
     for(e = ; e <= k[] ; e++)
     for(o =  ; o <= k[] ;o++)
     for(g =  ; g <= k[] ; g++)
     {
         dp[i][j][e][o][g] = i*p[]+j*p[]+e*p[]+o*p[]+g*p[];
         for(q = ; q <= s ; q++)
         {
             int i1=,i2=,i3=,i4=,i5=;
             for(a = ; a <= pp[q].n ;a++)
             {
                 if(pp[q].c[a]==c[])
                 i1 = pp[q].k[a];
                 else if(pp[q].c[a]==c[])
                 i2 = pp[q].k[a];
                 else if(pp[q].c[a]==c[])
                 i3 = pp[q].k[a];
                 else if(pp[q].c[a]==c[])
                 i4 = pp[q].k[a];
                 else
                 i5 = pp[q].k[a];
             }
             if(i-i1>=&&j-i2>=&&e-i3>=&&o-i4>=&&g-i5>=)
             {
                 dp[i][j][e][o][g] = min(dp[i][j][e][o][g],dp[i-i1][j-i2][e-i3][o-i4][g-i5]+pp[q].p);
             }
         }
     }
     cout<<dp[k[]][k[]][k[]][k[]][k[]]<<endl;
     return ;
 }