Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

题解：

裸的01背包，经典例题。

参考代码：

#include<stdio.h>
#define M 1111
#include<string.h>
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int t,a[M],b[M],dp[M];
	scanf("%d",&t);
	while(t--)
	{
		int n,v;
		memset(dp,0,sizeof(dp));
		scanf("%d%d",&n,&v);
		for(int i=0;i<n;i++)
		    scanf("%d",&b[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<n;i++)
        {
        	for(int j=v;j>=a[i];j--)
        	{
	        	dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
	        }
        }
        printf("%d\n",dp[v]);
	}
	return 0;
}

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