HDU 2602 Bone Collector

题目
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

题意：有一个人喜欢收集骨头，给你背包的体积，每种骨头的价值和体积，判断背包里可以装的最大价值。
直接套背包的公式就可以。

#include <stdio.h>
#include <algorithm>
using namespace std;
long long t,n,v,val[1010],vol[1010],dp[1010][1010];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&v);
		for(int i=1;i<=n;i++)
		    scanf("%d",&val[i]);
		for(int i=1;i<=n;i++)
		    scanf("%d",&vol[i]);
		for(int i=1;i<=n;i++)
		    for(int j=0;j<=v;j++)
		        if(vol[i]<=j) dp[i][j]=max(dp[i-1][j],dp[i-1][j-vol[i]]+val[i]);
		        else dp[i][j]=dp[i-1][j];
		printf("%lld\n",dp[n][v]);
	}
	return 0;
}