HDOJ-2602 Bone Collector [DP-01背包问题]

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12502    Accepted Submission(s): 4874


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
14
 

 

Author
Teddy
 

 

Source
 

 

Recommend
lcy
 
 
 
code:
 1 #include<iostream>
 2 using namespace std;
 3 
 4 int main()
 5 {
 6     int N,V;
 7     int n[1001],v[1001],dp[1001];
 8     int T;
 9     int i,j;
10     while(~scanf("%d",&T))
11     {
12         while(T--)
13         {
14             scanf("%d%d",&N,&V);
15             for(i=1;i<=N;i++)
16                 scanf("%d",&v[i]);
17             for(i=1;i<=N;i++)
18                 scanf("%d",&n[i]);
19  //------------------------------------------------------------------------------------------------------------- 
20             memset(dp,0,sizeof(dp));
21             for(i=1;i<=N;i++)
22                 for(j=V;j>=n[i];j--)
23                 {
24                     if(dp[j-n[i]]+v[i]>dp[j])
25                         dp[j]=dp[j-n[i]]+v[i];
26                 }
27  //------------------------------------------------------------------------------------------------------------- 
28             printf("%d\n",dp[V]);
29         }
30     }
31     return 0;
32 } 

 

 

自己调试结果:

最大容量V 物品个数N Bone Collector
10 5
物品大小n[i] 物品价值v[i] 编号 DP 10 9 8 7 6 5 4 3 2 1
5 1 1 1 1 1 1 1 1 1 0 0 0 0
4 2 2 2 3 3 2 2 2 2 2 0 0 0
3 3 3 3 6 5 5 5 3 3 3 3 0 0
2 4 4 4 9 9 7 7 7 7 7 4 4 0
1 5 5 5 14 12 12 12 12 12 9 9 5 5
posted @ 2012-07-18 12:13  max_xbw  阅读(312)  评论(0编辑  收藏  举报