bone collector hdu 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

 

先将原始问题一般化，欲求背包能够获得的总价值，即欲求前j个物体放入容量为m（kg）背包的最大价值f[j]——使用一个数组来存储最大价值，当j取10时，即原始问题了。而前i个物体放入容量为m（kg）的背包，又可以转化成前(i-1)个物体放入背包的问题。

核心代码：

for(i=0;i<n;i++)
            for(j=v;j>=bone[i].volume;j--)
            f[j]=max(f[j],f[j-bone[i].volume]+bone[i].value);

 f[j]=max(f[j],f[j-bone[i].volume]+bone[i].value);即为该问题的状态转移方程

当i==0,bone[0].volume==5时经过一个循环

f[10]=1;

f[9]=1;

f[8]=1;

f[7]=1;

f[6]=1;

f[5]=1;

f[4]=0;

f[3]=0;

f[2]=0;

f[1]=0;

f[0]=0;

当i==1,bone[1].volume==4时经过一个循环

f[10]=3;

f[9]=3;

f[8]=2;//因为飞f[4]==0，f[8]==1,而8-bone[1].volume==4，所以f[8]=max(f[8],f[j-bone[i].volume（4）]+bone[i].value（2）);下面同理

f[7]=2;

f[6]=2;

f[5]=2;

f[4]=2;

f[3]=0;//3<5也<4,所以前两个骨头都放不下，下面同理，上面也同理

f[2]=0;

f[1]=0;

f[0]=0;

当i==2,bone[2].volume==3时经过一个循环

f[10]=5;

f[9]=5;

f[8]=5;

f[7]=5;

f[6]=3;

f[5]=3;

f[4]=3;

f[3]=3;

f[2]=0;

f[1]=0;

f[0]=0;

当i==3,bone[3].volume==2时经过一个循环

f[10]=9;

f[9]=9;

f[8]=7;

f[7]=7;

f[6]=7;

f[5]=7;

f[4]=4;

f[3]=4;

f[2]=4;;//在此j循环跳出，上面同理，下面也同理，f[j]>=2;

f[1]=0;

f[0]=0;

当i==4,bone[3].volume==1时经过一个循环

f[10]=14;

f[9]=12;

f[8]=12;

f[7]=12;

f[6]=12;

f[5]=9;

f[4]=9;

f[3]=9;

f[2]=5;;//在此j循环跳出，上面同理，下面也同理，f[j]>=2;

f[1]=5;

f[0]=0;

 代码实现：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct bone
{
    int volume;
    int value;
}bone[1005];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int t,n,v,i,j,f[1005];
    while(cin>>t)
    {
        while(t--)
        {
            cin>>n>>v;
            for(i=0;i<n;i++)
            cin>>bone[i].value;
            for(i=0;i<n;i++)
            cin>>bone[i].volume;
            memset(f,0,sizeof(f));
            for(i=0;i<n;i++)
                for(j=v;j>=bone[i].volume;j--)
                    f[j]=max(f[j],f[j-bone[i].volume]+bone[i].value);
            cout<<f[v]<<endl;
        }
    }
    return 0;
}