Bone Collector（背包问题）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 105094    Accepted Submission(s): 42577

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 2 ³¹).

 

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

 

Recommend

lcy

 

Statistic |  Submit |  Discuss | Note

许多年前，在泰迪的家乡，有一个人被称为“骨头收藏家”。这个男人喜欢收集各种各样的骨头，比如狗的，牛的，还有他去了坟墓… “骨头收藏家”有一个体积为V的大袋子，沿着他的收集之旅有很多骨头，显然，不同的骨骼有不同的价值和不同的体积，现在根据他的行程给出每个骨头的值，你能计算超出骨收集器可以获得的总价值的最大值？

思路：
f(i,j)表示有i件物品，背包容量为j时，可以得到的最大价值。
那么f(i,j)=max{f(i-1,j),v(i)+f(i-1,j-w(i))}
其中v(i),w(i)分别表示第i件物品的价值和重量
这个式子的含义为：对于有i件物品，背包容量为j，可以得到的最大价值等于f(i-1,j)，即不取这件物品；或者v(i)+f(i-1,j-w(i))，即取这件物品，这两种可能的最大值。
再解释一下第二个：
v(i)+f(i-1,j-w(i))，第i件的物品的价值，再加上剩余背包容量在不考虑i物品的情况下能取到的最大值。

#include<iostream>
using namespace std;
const int MAX=1001;
int dp(int v[],int w[],int n,int bag){
    int arr[1001]={};           //这里把二维数组压缩成以为数组
    
    for(int i=1;i<=n;i++)
        for(int j=bag;j>=w[i];j--)      //从后往前计算，因为每一个位置的值依赖与当前位置的左边位置的值
                arr[j]=arr[j]>(v[i]+arr[j-w[i]])?arr[j]:(v[i]+arr[j-w[i]]); //f(i,j)=max{f(i-1,j),v(i)+f(i-1,j-w(i))}
    return arr[bag];
}
int main(){
    int t;
    cin>>t;
    for(int i=0;i<t;i++){
        int v[MAX],w[MAX];
        int n,bag;
        cin>>n>>bag;
        for(int i=1;i<=n;i++){
            cin>>v[i];
        }
        for(int i=1;i<=n;i++){
            cin>>w[i];
        }
        cout<<dp(v,w,n,bag)<<endl;
    }
    return 0;
}