Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40598 Accepted Submission(s): 16872

Problem Description

Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?

Input

The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.

Output

One integer per line representing the maximum of the total value (this number will be less than 2³¹).

Sample Input

1 5 10 1 2 3 4 5 5 4 3 2 1

Sample Output

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

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 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<stack>
 5 #include<set>
 6 #include<map>
 7 #include<queue>
 8 #include<algorithm>
 9 using namespace std;
10 #define max(a,b) (a>b?a:b)
11 int va[1005],vo[1005],dp[1005][1005];
12 int main()
13 {
14     //freopen("D:\\INPUT.txt","r",stdin);
15     int t,i,j;
16     int n,v;
17     scanf("%d",&t);
18     while(t--)
19     {
20         scanf("%d %d",&n,&v);
21         for(i=1; i<=n; i++)
22         {
23             scanf("%d",&va[i]);
24         }
25         for(i=1; i<=n; i++)
26         {
27             scanf("%d",&vo[i]);
28         }
29         //dp[i][j]  前i件物品放入j体积的价值的最大值
30         //dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i])
31         for(i=1; i<=n; i++) //i体积
32         {
33             for(j=0; j<=v; j++)
34             {
35                 if(j>=vo[i]){
36                     dp[i][j]=max(dp[i-1][j],dp[i-1][j-vo[i]]+va[i]);
37                 }
38                 else{
39                     dp[i][j]=dp[i-1][j];
40                 }
41
42                 //cout<<i<<"  "<<j<<" "<<dp[i][j]<<endl;
43             }
44         }
45         printf("%d\n",dp[n][v]);
46     }
47     return 0;
48 }

转载于:https://www.cnblogs.com/Deribs4/p/4796120.html

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