HDU 2602 Bone Collector

出了套题，发现自己已经不会写背包了。。标程都拿不出手来。。。找些题练练手，特意开了个背包分类。

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 61370    Accepted Submission(s): 25570

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

 

这个题不需要什么题目大意和思路。。01背包裸题，不过我是第一次写一维的数组，没注意到第二层循环j需要反着从大往小跑。

附代码。

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<cmath>#include<cstdlib>#include<ctime>using namespace std;const int MAXN=4000;int w[MAXN],v[MAXN];int f[13000];void work1(int n,int m){memset(f,0,sizeof(f));for (int i=1;i<=n;i++){for (int j=m;j>=v[i];j--){f[j]=max(f[j],f[j-v[i]]+w[i]);}}printf("%d\n",f[m]);}int main(){int n,m,t;scanf("%d",&t);while (t){scanf("%d%d",&n,&m);for (int i=1;i<=n;i++){scanf("%d",&w[i]);}for (int i=1;i<=n;i++){scanf("%d",&v[i]);}t--;work1(n,m);}     return 0;}

MAX随便设的不要在意这种细节。。。