HDU 2602 Bone Collector(01背包)

Posted 2020-06-13

Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 44243    Accepted Submission(s): 18434     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     Author Teddy     Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 int dp[1010];
 6 struct node
 7 {
 8     int value;
 9     int volume;
10 }m[1010];
11 int main()
12 {
13     int t;
14     cin>>t;
15     int n,v;
16     while(t--){
17         cin>>n>>v;
18         memset(dp,0,sizeof(dp));
19         for(int i=1;i<=n;i++) cin>>m[i].value;
20         for(int i=1;i<=n;i++) cin>>m[i].volume;
21         for(int i=1;i<=n;i++){
22             for(int j=v;j>=m[i].volume;j--){
23                 dp[j]=max(dp[j],dp[j-m[i].volume]+m[i].value);
24             }
25         }
26         cout<<dp[v]<<endl;
27     }    
28     return 0;
29 }