M

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

分析：就是让你求怎样装东西能够在背包可以承受的重量内所装的物品价值最高

问题实质0-1背包问题，恩，还不是很理解，怎么说呢，做出来不难，但是理解的话就有问题了；
AC代码：

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int d[1001];
int va[1001],vo[1001];
int max(int a,int b)
{
    return a>b?a:b;
}
int main()
{
    int t,m,n;
    scanf("%d",&t);
    while(t--)
    {
        memset(d,0,sizeof(d));
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; i++)
            scanf("%d",&va[i]);
        for(int i=1; i<=n; i++)
            scanf("%d",&vo[i]);
        for(int i=1; i<=n; i++)
        {
            for(int j=m; j>=vo[i]; j--)
            {
                d[j]=max(d[j],d[j-vo[i]]+va[i]);
//Dp状态方程dp[i] = max(dp[i],dp[i - wi] + vi) 
//dp[i]表示容量为i时获得的头骨的最大价值
            }
        }
        printf("%d
",d[m]);
    }
    return 0;
}