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。

 1 /*
 2     ID: shangca2
 3     LANG: C++
 4     TASK: shopping
 5  */
 6 #include <iostream>
 7 #include<cstdio>
 8 #include<cstring>
 9 #include<algorithm>
10 #include<stdlib.h>
11 using namespace std;
12 #define INF 0xfffffff
13 int dp[6][6][6][6][6];
14 struct node
15 {
16     int c[6],k[6],p,n;
17 }pp[110];
18 int c[6],k[6],p[6];
19 int main()
20 {
21     freopen("shopping.in","r",stdin);
22     freopen("shopping.out","w",stdout);
23     int i,j,s,b,e,o,g,q,a;
24     for(i =0 ; i <= 5 ; i++)
25         for(j = 0 ; j <= 5 ; j++)
26             for(e = 0 ; e <= 5 ; e++)
27                 for(o = 0 ; o <= 5 ; o++)
28                     for(g = 0 ; g <= 5 ; g++)
29                     dp[i][j][e][o][g] = INF;
30     cin>>s;
31     for(i = 1; i <= s ; i++)
32     {
33         cin>>pp[i].n;
34         for(j = 1; j <= pp[i].n ; j++)
35         cin>>pp[i].c[j]>>pp[i].k[j];
36         cin>>pp[i].p;
37     }
38     cin>>b;
39     for(i =  1; i <= b ;i++)
40     cin>>c[i]>>k[i]>>p[i];
41     for(i = 0 ;i <= k[1] ; i++)
42     for(j = 0; j <= k[2] ; j++)
43     for(e = 0; e <= k[3] ; e++)
44     for(o = 0 ; o <= k[4] ;o++)
45     for(g = 0 ; g <= k[5] ; g++)
46     {
47         dp[i][j][e][o][g] = i*p[1]+j*p[2]+e*p[3]+o*p[4]+g*p[5];
48         for(q = 1; q <= s ; q++)
49         {
50             int i1=0,i2=0,i3=0,i4=0,i5=0;
51             for(a = 1; a <= pp[q].n ;a++)
52             {
53                 if(pp[q].c[a]==c[1])
54                 i1 = pp[q].k[a];
55                 else if(pp[q].c[a]==c[2])
56                 i2 = pp[q].k[a];
57                 else if(pp[q].c[a]==c[3])
58                 i3 = pp[q].k[a];
59                 else if(pp[q].c[a]==c[4])
60                 i4 = pp[q].k[a];
61                 else
62                 i5 = pp[q].k[a];
63             }
64             if(i-i1>=0&&j-i2>=0&&e-i3>=0&&o-i4>=0&&g-i5>=0)
65             {
66                 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);
67             }
68         }
69     }
70     cout<<dp[k[1]][k[2]][k[3]][k[4]][k[5]]<<endl;
71     return 0;
72 }
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posted @ 2013-08-21 00:10  _雨  阅读(232)  评论(0编辑  收藏  举报