【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 2 ³¹).

 

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

 

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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