NOIP2006 Jin Ming's Budget Plan

Jin Ming's budget plan

 

Obviously a knapsack problem

Put each main component and its corresponding accessories in a group, and enumerate each group. There are the following options:

1. do not choose

2. Only select the main part

3. One main part + one accessory

4. One main part + two accessories

 

So it became the 01 backpack. .

 

1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4  using  namespace std;
 5  int n,m,f[ 32010 ],v[ 65 ][ 4 ],w[ 65 ][ 4 ], pos[ 65 ]; // pos[i] records the subscript of the item number i in the array v, w 
　　　　　　　　　　　　　　　　　　　　　　　　　　　 //w[i][0] represents the number of attachments of the i-th main piece 

6  int main()
 7  {
 8      scanf( " %d%d " ,&n,& m);
 9      int V,P,Q,sum;
 10     for(int i=1;i<=m;i++)
11     {
12         scanf("%d%d%d",&V,&P,&Q);
13         if(Q==0)
14         {
15             v[++sum][1]=V;
16             w[sum][1]=P;
17             pos[i]=sum;
18         }
19         else
20         {
21             w[pos[Q]][++w[pos[Q]][0]+1]=P;
22             v[pos[Q]][w[pos[Q]][0]+1]=V;
23         }
24     }
25     for(int i=1;i<=sum;i++)
26      for(int j=n;j>=v[i][1];j--)
27      {
28          f[j]=max(f[j],f[j-v[i][1]]+w[i][1]*v[i][1]);
29          if(w[i][0]>=1&&j-v[i][1]-v[i][2]>=0)
30           f[j]=max(f[j],f[j-v[i][1]-v[i][2]]+w[i][1]*v[i][1]+w[i][2]*v[i][2]);
31          if(w[i][0]==2&&j-v[i][1]-v[i][3]>=0)
32          {
33               f[j]=max(f[j],f[j-v[i][1]-v[i][3]]+w[i][1]*v[i][1]+w[i][3]*v[i][3]);
34               if(j-v[i][1]-v[i][2]-v[i][3]>=0)
35                f[j]=max(f[j],f[j-v[i][1]-v[i][2]-v[i][3]]+w[i][1]*v[i][1]+w[i][2]*v[i][2]+w[i][3]*v[i][3]);
36         }
37      }
38     printf("%d\n",f[n]);
39     return 0;
40 }