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 2 ³¹).

 

 

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 }