hdu 2602 Bone Collector



A - Bone Collector

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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
     
     


     
     

      题意：一个体积为V的背包，来装不同的骨头，每个骨头有自己的体积和价值，求出这个背包装骨头的最大价值
     
     

      是01背包的的基础题目直接套用01背包的转移方程就可以AC
     
     

       二维数组dp
     
     

      
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int dp[1010][1010];
int value[1010];
int volume[1010];
int main()
{
    int t,i,j;
    cin>>t;
    int n,V;
    while(t--)
    {
        cin>>n;
        cin>>V;
        for(i=1; i<=n; i++)
            cin>>value[i];
        for(i=1; i<=n; i++)
            cin>>volume[i];
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; i++)
        {
            for(j=0; j<=V; j++)
            {
                if(volume[i]<=j&&dp[i-1][j]<dp[i-1][j-volume[i]]+value[i])
                    dp[i][j]=dp[i-1][j-volume[i]]+value[i];
                else
                    dp[i][j]=dp[i-1][j];
            }
        }
        cout<<dp[n][V]<<endl;
    }
    return 0;
}
      

     
     

      

     
     

      

     
     

      一维数组dp
     
     

      
#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int main()
{
    int dp[10000];
    int t;
    int v[10000] , val[10000];
    int n , V;
    cin>>t;
    while(t--)
    {
        cin>>n>>V;
        for(int i=1;i<=n;i++)
            cin>>val[i];
        for(int i=1;i<=n;i++)
            cin>>v[i];
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            for(int j=V;j>=v[i];j--)
                dp[j]=max(dp[j-v[i]]+val[i],dp[j]);
            cout<<dp[V]<<endl;
    }
    return 0;
}