HDU 2602 Bone Collector（ 01揹包 ）

一道簡單的01揹包問題

Bone Collector

                                                       Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                       Total Submission(s): 18975    Accepted Submission(s): 7510

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

 

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;
int cas;
int n,m;
int dp[1005],v[1005],w[1005];

int main()
{
	scanf("%d",&cas);
	while( cas-- ){
		scanf("%d%d",&n,&m);
		memset( dp , 0 , sizeof(dp ));
		for( int i=1 ; i<=n ; i++ )
			scanf("%d",&w[i]);
		for( int i=1 ; i<=n ; i++ )
			scanf("%d",&v[i]);
		for( int i=1 ; i<=n ; i++ ){
			for( int V=m ; V>=v[i] ; V-- ){
				dp[V] = max( dp[V] , dp[V-v[i]] + w[i] );	
			}	
		}
		printf("%d\n",dp[m]);
	}
	return 0;
}