模板--完全背包HDU--2602 Bone Collector

Posted 2020-10-02 ⊙∽⊙Perseverance

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

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 long long dp[1000000+10];
 5 int w[1100],p[1100];
 6 int main()
 7 {
 8     int t,N,W;
 9     scanf("%d",&t);
10     while(t--)
11     {
12         scanf("%d %d",&N,&W);
13         for(int i=1;i<=N;i++)   //
14             scanf("%d",&p[i]);
15         for(int i=1;i<=N;i++)   //
16             scanf("%d",&w[i]);    
17         for(int i=0;i<=W;i++)
18             dp[i]=0;
19         for(int i=1;i<=N;i++)  // 这几个地方i开头值注意一样  错了几次..
20         {
21             for(int j=W;j>=w[i];j--)
22             {
23                 dp[j]=max(dp[j],dp[j-w[i]]+p[i]);  //核心代码
24             }    
25         }
26         printf("%lld\n",dp[W]);            
27         
28     }
29     
30     return 0;
31 }