poj 1041Dividing解题报告可行性背包问题

题意是说，给一些大理石，每种有num[i]个，判断能否分成两部分，可行性问题，感觉不像是动态规划，中间没有最优子结构这个性质，倒像是纯粹的递推

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 1 #include<stdio.h>
 2 #include<string.h>
 3 #define N 60005
 4 int bag[7];
 5 int v[N];
 6 bool dp[N];
 7 int main()
 8 {
 9     int sum;
10     int i,j,k,ans;
11     int icase=1;
12     while(1)
13     {
14         sum=0;
15         for(i=1;i<=6;i++)
16         {
17             scanf("%d",&bag[i]);
18             sum+=bag[i]*i;
19         }
20         if(!sum)
21         break;
22         printf("Collection #%d:\n",icase++);
23         if(sum&1)
24         {
25             printf("Can't be divided.\n\n");
26             continue;
27         }
28         memset(dp,0,sizeof(dp));
29         ans=sum>>1;
30         k=0;
31         for(i=1;i<=6;i++)
32         {
33             j=1;
34             while(j<=bag[i])
35             {
36                 bag[i]-=j;
37                 v[k++]=j*i;
38                 j<<=1;
39             }
40             if(bag[i])
41             v[k++]=bag[i]*i;
42         }
43         dp[0]=true;
44         for(i=0;i<k;i++)
45         for(j=ans;j>=v[i];j--)
46         if(dp[j-v[i]])
47         dp[j]=true;
48         if(dp[ans])
49         printf("Can be divided.\n\n");
50         else
51         printf("Can't be divided.\n\n");
52     }
53     return 0;
54 }