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): 40656    Accepted Submission(s): 16898

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

15 101 2 3 4 55 4 3 2 1

 

Sample Output

14

 

Author

Teddy

嗯，题目大意就是说，背包容量为V然后不同的骨头有不同的体积和价值，问在背包容量范围内所能装的骨头的最大总价值

注意数据范围

#include <iostream>#include<cstdio>#include<cstring>#define maxn 1001using namespace std;__int64 c[maxn],w[maxn],f[maxn];__int64 max(__int64 a,__int64 b){    return a>b?a:b;}int main(){    __int64 t,n,v;    scanf("%I64d",&t);    while(t--)    {        scanf("%I64d%I64d",&n,&v);        for(int i=1;i<=n;++i)            scanf("%I64d",&w[i]);        for(int i=1;i<=n;++i)            scanf("%I64d",&c[i]);        memset(f,0,sizeof(f));        for(int i=1;i<=n;++i)        {            for(int j=v;j>=c[i];j--)            {                f[j]=max(f[j],f[j-c[i]]+w[i]);            }        }        printf("%I64d\n",f[v]);    }    return 0;}