HDU2602Bone Collector 简单0-1背包

Posted 2020-07-25 asuml

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 48618    Accepted Submission(s): 20269

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

注意：先输入每个骨头的价值然后输入每个骨头的体积。

状态转移方程：dp[j]=max(dp[j],dp[j-v[i]]+w[i]。

#include <iostream>
#include <string.h>

using namespace std;
int N,V;
int dp[1111];
int w[1111],v[1111];
int main()
{
    int t;
    while(cin>>t){
        while(t--){
            cin>>N>>V;
            for(int i=1;i<=N;i++){
                cin>>w[i];
            }
            for(int i=1;i<=N;i++){
                cin>>v[i];
            }
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=N;i++){
                for(int j=V;j>=v[i];j--){       //j如果小于了v[i]那么v[i]一定无法装入袋子
                    dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
                }
            }
            cout<<dp[V]<<endl;
        }
    }
    return 0;
}