HDU2602——Bone Collector【01背包问题】

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

 

 

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01背包模版题。

#include
#include
#include
#include
using namespace std;
const int MAXN=1010;
int dp[MAXN],w[MAXN],v[MAXN];//容量对应的最大价值，单个物品的价值，物品的体积
int main(){
    int t;
    cin>>t;
    while(t--){
        int n,s;
        cin>>n>>s;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++) cin>>w[i];
        for(int i=1;i<=n;i++) cin>>v[i];
        for(int i=1;i<=n;i++){
            for(int j=s;j>=v[i];j--){
                dp[j]=max(dp[j],dp[j-v[i]]+w[i]);
            }
        }
        printf("%d\n",dp[s]);
    }
    return 0;
}