hdu 2602 Bone Collector （简单01背包）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30486    Accepted Submission(s): 12550

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

 

 

题目大意：在所给的体积范围内，拿到尽可能多的价值的骨头。

注意：先给的是每个骨头的价值，再给的是每个骨头的体积。因为弄反而wa就很不值了哦~

 

详见代码。

 1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <cstring> 5 using namespace std; 6  7 int main () 8 { 9     int t;10     int dp[1010],v[1010],w[1010];11 12     while (cin>>t)13     {14         while (t--)15         {16             memset(dp,0,sizeof(dp));17             int n,V,i,j;18             cin>>n>>V;19             for ( i=1; i<=n; i++)20                 cin>>w[i];21             for (i=1; i<=n; i++)22                 cin>>v[i];23             for (i=1; i<=n; i++)24             {25                 for (j=V; j>=v[i]; j--)26                 {27                     dp[j]=max(dp[j],dp[j-v[i]]+w[i]);28                     //cout<<dp[j]<<" "<<v[i]<<endl;29                 }30             }31             printf ("%d\n",dp[V]);32         }33 34     }35     return 0;36 }