hdu Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 226    Accepted Submission(s): 115

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

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

Recommend

lcy

 

分析： 典型的0-1背包，V不要求用完。

#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<iostream>
using namespace std;
typedef struct S
{
    int v,w;
}BONE;
BONE bone[1010];

int f[1010];
int main()
{
    int T,W,n,i,j,stop;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&W);
        for(i=0;i<n;++i)
            scanf("%d",&bone[i].v);
        for(i=0;i<n;++i)
            scanf("%d",&bone[i].w);
        memset(f,0,sizeof(f));
       for(i=0;i<n;++i)
       {
           for(j=W;j>=bone[i].w;--j)
           {
               if(f[j-bone[i].w]+bone[i].v>f[j])
                   f[j]=f[j-bone[i].w]+bone[i].v;
           }
       }
        printf("%d\n",f[W]);
    }
    return 0;
}