hdu2602 Bone Collector 01背包

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

 

题解：

      在容量为m的背包中放入骨头，每个骨头有一定的体积和一定的价值，问如何装能使背包中背的骨头价值最大。典型的01背包问题，套用模板即可。

      我写了两个，分别是一维数组和二维数组，一起发上来。

 

代码1：

#include <bits/stdc++.h>

using namespace std;
int t,n,m,v[],w[],i,j,f[][];
int main()
{
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(i=;i<=n;i++) cin>>v[i];
        for(i=;i<=n;i++) cin>>w[i];
        memset(f,,sizeof(f));
        for(i=;i<=n;i++)          //装入第i个骨头时，背包容量为j时的最大价值
        for(j=;j<=m;j++){
            if(j<w[i]){
                f[i][j]=f[i-][j];
            }
            else{
                f[i][j]=max(f[i-][j],f[i-][j-w[i]]+v[i]);
            }
        }
        cout<<f[n][m]<<endl;
    }
    return ;
}

代码2：

#include <bits/stdc++.h>

using namespace std;
int n,m,f[],v[],w[],i,j,t;

int main()
{
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(i=;i<=n;i++) cin>>v[i];
        for(i=;i<=n;i++) cin>>w[i];
        memset(f,,sizeof(f));
        for(i=;i<=n;i++)
        for(j=m;j>=w[i];j--){
            if(f[j]<f[j-w[i]]+v[i])
                f[j]=f[j-w[i]]+v[i];
        }
        cout<<f[m]<<endl;
    }
    return ;
}