hdu2602-01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 23108    Accepted Submission(s): 9379

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

 

 

二维数组

#include <iostream>
#include <cstring>
#define maxn 1005
using namespace std;
int n,c[maxn],v[maxn],ans[maxn][maxn],T,V;

int main()
{
    cin>>T;
    while(T--){
        cin>>n>>V;
        for(int i=1; i<=n; i++)
            cin>>v[i];
        for(int i=1; i<=n; i++)
            cin>>c[i];
        memset(ans,0,sizeof(ans));
        for(int i=1; i<=n; i++)
            for(int j=0; j<=V; j++){   //j要从0开始，从1开始会Wa，目前还不是很理解这点
                if(j<c[i]) 
                    ans[i][j]=ans[i-1][j];
                else 
                    ans[i][j]=max(ans[i-1][j], ans[i-1][j-c[i]]+v[i]);
            }
        cout<<ans[n][V]<<endl;
    }
    return 0;
}

 

一维数组

#include <iostream>
#include <cstring>
#define maxn 1005
using namespace std;

int N,V,ans[maxn],c[maxn],w[maxn];

void ZeroOnePack(int c, int w)
{
    for(int i=V; i>=c; i--)
        ans[i]=max(w+ans[i-c],ans[i]);
}
int main()
{
    int T;
    cin>>T;
    while(T--){
        cin>>N>>V;
        memset(ans,0,sizeof(ans));
        for(int i=1; i<=N; i++)
            cin>>w[i];
        for(int i=1; i<=N; i++)
            cin>>c[i];
        for(int i=1; i<=N; i++)
            ZeroOnePack(c[i],w[i]);
        cout<<ans[V]<<endl;
    }
    return 0;
}