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
/*题解：
01背包：状态转移方程为：f[i][j] = max(f[i - 1][j], f[i - 1][j - c[i]] + w[i]);
*/
#include<stdio.h>
#include<queue>
#include<iostream>
#include<string.h>
using namespace std;
int f[1005][1005];
int main()
{
int w[1005], c[1005];
int n, v;
int T;
cin >> T;
while (T--)
{
cin >> n >> v;
memset(f, 0, sizeof(f));
memset(w, 0, sizeof(w));
memset(c, 0, sizeof(c));
for (int i = 1; i <= n; i++)
{
cin >> w[i];//价值
}
for (int i = 1; i <= n; i++)
cin >> c[i];//体积
for (int i = 1; i <= n; i++)
{
for (int j = 0; j <= v; j++)
{
if (j >= c[i])
{
f[i][j] = max(f[i - 1][j], f[i - 1][j - c[i]] + w[i]);
}
else
f[i][j] = f[i - 1][j];
}
}
cout << f[n][v] << endl;
}
return 0;
}
