Bone Collector HDU - 2602

标签：背包   bsp   img   ase   iostream   cti   cow   call   set   

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 ?

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

裸的01背包

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 using namespace std;
 6 
 7 int main()
 8 {   int t,n,V;
 9     cin>>t;
10     while(t--){
11         scanf("%d%d",&n,&V);
12         int va[1005],vo[1005],dp[1005];
13         for(int i=0;i<n;i++) scanf("%d",&va[i]);
14         for(int i=0;i<n;i++) scanf("%d",&vo[i]);
15         
16         memset(dp,0,sizeof(dp));
17         for(int i=0;i<n;i++){
18             for(int j=V;j>=vo[i];j--)
19                 dp[j]=max(dp[j],dp[j-vo[i]]+va[i]);
20         }
21         printf("%d\n",dp[V]);
22     }
23 }

Bone Collector HDU - 2602

标签：背包   bsp   img   ase   iostream   cti   cow   call   set   

原文地址：http://www.cnblogs.com/zgglj-com/p/6810482.html