HDU 2602 Bone Collector

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

Author
Teddy
 
Source
HDU 1st “Vegetable-Birds Cup” Programming Open Contest

Problem analysis

这是一个0-1背包问题，每个骨头只取一次（我觉得这篇文章解释0-1背包问题解释的非常清楚，有不懂得可以去看一下）。

然后就是要注意骨头的体积可以取零！！！

code implementation

#include <bits/stdc++.h>
using namespace std;
int w[1001],a[1001],dp[1001][1001];
int main()
{
    int n,v,k;
    cin>>k;
    while(k--)
    {
        memset(dp,0,sizeof(dp));
        cin>>n>>v;
        for(int i=1;i<=n;i++)
        {
            cin>>w[i];//价值 
        }
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];    //体积 
        }
        
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<=v;j++)//这里一定要用j=0，虽然我也不知道为啥骨头的体积可以为零，都没了还捡啥 
            {
                if(j>=a[i])
                {
                    dp[i][j]=max(dp[i-1][j],dp[i-1][j-a[i]]+w[i]);//i-1!!
                }
                else{
                    dp[i][j]=dp[i-1][j];
                }
            }
        }
        cout<<dp[n][v]<<endl;
    }
    return 0;
}