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
#include<stdio.h>
#include<string.h>
int dp[1010][1010];
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
int t,n,c,i,j;
int w[1000],v[1000];
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&c);
for(i=1;i<=n;i++)
scanf("%d",&w[i]);
for(i=1;i<=n;i++)
scanf("%d",&v[i]);
memset(dp,0,sizeof(dp));
for(i=1;i<=n;i++)
{
for(j=0;j<=c;j++)
{
if(v[i]<=j)//表示第i个物品将放入大小为j的背包中
dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i]]+w[i]);//第i个物品放入后，那么前i-1个物品可能会放入也可能因为剩余空间不够无法放入
else //第i个物品无法放入
dp[i][j]=dp[i-1][j];
}
}
printf("%d\n",dp[n][c]);
}
return 0;
}
