【杭电0j2602】Bone Collector

                                                 Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51528    Accepted Submission(s): 21697

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

 

Source

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

 

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#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 1002;int val[N],vol[N],dp[N];int main() {int T;scanf("%d",&T);while(T--) {int n,V;memset(dp,0,sizeof(dp));scanf("%d%d",&n,&V);for(int i=1; i<=n; i++) scanf("%d",&val[i]);for(int j=1; j<=n; j++) scanf("%d",&vol[j]);for(int i=1; i<=n; i++) {for(int j=V; j>=vol[i]; j--) {//倒序，关键 if(dp[j-vol[i]]+val[i]>dp[j]) {//正序的话会使上一个i被忽略 dp[j]=dp[j-vol[i]]+val[i];}}}printf("%d\n",dp[V]);}return 0;}

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2602