HDU 2602

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：输入N,V。N表示有多少骨头。V表示你的背包有多大。然后输入序列表示每个骨头的价值，再输入序列表示每个骨头的体积。

　　　　要求输出  背包装骨头可以装的最大的价值是多少....

 

 

 

思路：动态规划

 

     如果可以装进背包，就把符合条件且价值最大的装进去

　　　不可以就不装.....

　　　这里用d[i][j]表示装第i个骨头时背包里的骨头价值，i表示第几个骨头，j表示背包里的体积.....

 

 

 

　　　　代码如下：

 

 

#include <stdio.h>
#include <string.h>
int n[1005],v[1005],d[1005][1005];
int max(int x,int y)
{
    return x>y?x:y;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int N,V;
        scanf("%d%d",&N,&V);
        for(int i=1;i<=N;i++)
            scanf("%d",&n[i]);
        for(int j=1;j<=N;j++)
            scanf("%d",&v[j]);
        for(int i=1;i<=N;i++)
        {
            for(int j=V;j>=0;j--)
            {
                if(j>=v[i])
                    d[i][j]=max(d[i-1][j],d[i-1][j-v[i]]+n[i]);
                else
                    d[i][j]=d[i-1][j];
            }
        }
        printf("%d\n",d[N][V]);

    }
}