3017

3017

Problem Q

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 62   Accepted Submission(s) : 22

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 …<br>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 ?<br><center><img src=../../../data/images/C154-1003-1.jpg> </center><br>

 

Input

The first line contain a integer T , the number of cases.<br>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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

题意：

不同的骨头有不同的价值，在袋子的体积确定的情况下求收藏家所能得到的最大价值；

思路：

用f[i][j]表示前i件物品用最大负重j的袋子所能装的最大权值

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]};

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int w[1005];
int v[1005];
int dp[1005];
int main()
{
    int t,n,m;
    scanf("%d",&t);
    while(t--)
    {
        memset(w,0,sizeof(w));
        memset(v,0,sizeof(v));
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&v[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&w[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=0;j--)
            {
                if(j>=w[i])//注意=号
                dp[j]=max(dp[j],dp[j-w[i]]+v[i]);
            }
        }
        printf("%d\n",dp[m]);
    }
    return 0;
}