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 ? 
InputThe 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.OutputOne integer per line representing the maximum of the total value (this number will be less than 2 31).Sample Input
1 5 10 1 2 3 4 5 5 4 3 2 1
Sample Output
14
1 #include<bits/stdc++.h>
2 using namespace std;
3 int main()
4 {
5 int dp[1005],n,v,nv[1005],nm[1005],t;
6 while(cin>>t)
7 {
8 while(t--)
9 {
10 memset(dp,0,sizeof(dp)); /*初始化*/
11 scanf("%d %d",&n,&v); /*读取骨头数量和背包重量*/
12 for(int i = 0; i < n; i++) /*每一个骨头的价值*/
13 scanf("%d",&nv[i]);
14 for(int i = 0; i < n; i++) /*每个骨头的重量*/
15 scanf("%d",&nm[i]);
16
17 for(int i = 0; i < n; i++)
18 for(int j = v; j >= nm[i];j--) /*dp算法*/
19 dp[j] = max(dp[j],dp[j-nm[i]]+nv[i]);
20 cout<<dp[v]<<endl;
21 }
22 }
23
24 return 0;
25 }
参考博客：
http://www.acmerblog.com/hdu-2602-bone-collector-4159.html
http://blog.csdn.net/someday7_toi/article/details/7781129
http://www.importnew.com/13072.html
http://blog.sina.com.cn/s/blog_8cf6e8d90100zldn.html