HDU 2602 Bone Collector (01背包)

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 <cstdio>
#include <cstring>
#define N 1010
int v[N], w[N];
int dp[N][N];
int n, m;
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		scanf("%d%d", &n, &m);
		for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
		for (int i = 1; i <= n; i++) scanf("%d", &v[i]);
		memset(dp, 0, sizeof(dp));
		for (int i = 1; i <= n; i++)
			for (int j = 0; j <= m; j++)
			{
				if (j >= v[i] && w[i] + dp[i - 1][j - v[i]] > dp[i - 1][j])
					dp[i][j] = w[i] + dp[i - 1][j - v[i]];
				else dp[i][j] = dp[i - 1][j];
			}
		printf("%d\n", dp[n][m]);
	}
	return 0;
}