hdu 2602 Bone Collector

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

这是一道简单的01背包的模板题。不过在做的时候，竟然把体积与价值弄混了，这也说明了我英语的不行，有待加强啊

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>	
using namespace std;
struct node
{
	int cost, value;
}a[1005];
int dp[1005];
int main()
{
	__int64  t, m, n, i,j;
	cin >> t;
	while (t--)
	{
		cin >> n >> m;
		for (i = 1;i <= n;i++)
			cin >> a[i].value;
		for (i = 1;i <= n;i++)
			cin >> a[i].cost;
		memset(dp, 0, sizeof(dp));
		for (i = 1;i <= n;i++)
		{
			for (j = m;j >= a[i].cost;j--)
			{
				dp[j] = max(dp[j], dp[j - a[i].cost] + a[i].value);
			}
		}
		cout << dp[m] << endl;
	}
	return 0;
}