ACM Bone Collector

 

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

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int dp[1005],n,v,nv[1005],nm[1005],t;
    while(cin>>t)
    {
        while(t--)
        {
            memset(dp,0,sizeof(dp));  /*初始化*/
            scanf("%d %d",&n,&v);         /*读取骨头数量和背包重量*/
            for(int i = 0; i < n; i++)  /*每一个骨头的价值*/ 
                scanf("%d",&nv[i]);
            for(int i = 0; i < n; i++)    /*每个骨头的重量*/ 
                scanf("%d",&nm[i]);
                
            for(int i = 0; i < n; i++)
                for(int j = v; j >= nm[i];j--)  /*dp算法*/ 
                    dp[j] = max(dp[j],dp[j-nm[i]]+nv[i]);
            cout<<dp[v]<<endl;
        }
    }

 return 0;
}

 

 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