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

 

01背包问题，主要是状态转移方程的使用，可以参考一下百度文库的背包九讲，讲的非常清楚

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct node
{
    int jz;
    int tj;
}Node[1010];

int main()
{
    int t;
    cin>>t;
    int dp[1010];
    while(t--)
    {
        int n,v;
        cin>>n>>v;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
        {
            cin>>Node[i].jz;
        }
        for(int i=0;i<n;i++)
        {
            cin>>Node[i].tj;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=v;j>=Node[i].tj;j--)
            {
                dp[j]=max(dp[j],dp[j-Node[i].tj]+Node[i].jz);
            }
        }
        cout<<dp[v]<<endl;
    }
    return 0;
}