杭电2602

哎，动态规划背包搞了好多天，终于搞懂了。。。坚持就是胜利

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 43080    Accepted Submission(s): 17962

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
   
   
   
   

 

#include<iostream>
using namespace std;
#define MAXN 1000
int d[MAXN][MAXN];
int main()
{
	int t,n,m,a[MAXN],b[MAXN],i,j;
	while(cin>>t)
	{
		while(t--)
		{
			cin>>n>>m;
			for(i=1;i<=n;i++)
			cin>>a[i];
			for(j=1;j<=n;j++)
			cin>>b[j];
			for(i=1;i<=n;i++)
			{
				for(j=0;j<=m;j++)
				{
					if(j>=b[i])         //第i个物品装入背包 
					{
						d[i][j]=d[i-1][j]>(d[i-1][j-b[i]]+a[i])?d[i-1][j]:(d[i-1][j-b[i]]+a[i]);  
					}
					else        //第i个物品不装入 
					d[i][j]=d[i-1][j];
				}
			}
			cout<<d[n][m]<<endl;
		}
	}
	return 0;
}