ACM Bone Collector

Posted 如有一味绝境,非历十方生死

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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 ? 
技术分享
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 }

 

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