【模板--完全背包】HDU--2602 Bone Collector

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

 #include<cstdio>
 #include<algorithm>
 using namespace std;
 long long dp[+];
 int w[],p[];
 int main()
 {
     int t,N,W;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d %d",&N,&W);
         for(int i=;i<=N;i++)   //
             scanf("%d",&p[i]);
         for(int i=;i<=N;i++)   //
             scanf("%d",&w[i]);
         for(int i=;i<=W;i++)
             dp[i]=;
         for(int i=;i<=N;i++)  // 这几个地方i开头值注意一样  错了几次..
         {
             for(int j=W;j>=w[i];j--)
             {
                 dp[j]=max(dp[j],dp[j-w[i]]+p[i]);  //核心代码
             }
         }
         printf("%lld\n",dp[W]);            

     }

     return ;
 }