HDU 2602 Bone Collector (01背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30923    Accepted Submission(s): 12722

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

大致题意：有一个骨头收藏家，他喜欢收藏一些有价值的骨头，但是每个骨头是有价值的。先输入一组测试数据T，魅族测试数据第一行两个数n和m，代表有n个骨头和最大重量为m，接下来n行每行输入一个骨头的价值和重量。求不超过总量的最大价值是多少？

思路：还是套用模版，两层for循环，使用dp公式，dp[j]=dp[j-v[i]]+w[i](当j小于最大重量时，可以装多少价值的骨头)

#include<iostream>
using namespace std;
int w[1010],v[1010],f[1010];
int main()
{
    int T,n,m,i,j;
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(i=1;i<=n;i++)
            cin>>w[i];
        for(i=1;i<=n;i++)
            cin>>v[i];
        memset(f,0,sizeof(f));
        for(i=1;i<=n;i++)
        {
            for(j=m;j>=v[i];j--)
            {
                if(f[j-v[i]]+w[i]>f[j])
                    f[j]=f[j-v[i]]+w[i];
            }
        }
        cout<< f[m] <<endl;
          
    }
    return 0;
}