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 ?
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 ?
InputThe 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.OutputOne 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<bits/stdc++.h>
using namespace std;
int main()
{
int dp[],n,v,nv[],nm[],t;
while(cin>>t)
{
while(t--)
{
memset(dp,,sizeof(dp)); /*初始化*/
scanf("%d %d",&n,&v); /*读取骨头数量和背包重量*/
for(int i = ; i < n; i++) /*每一个骨头的价值*/
scanf("%d",&nv[i]);
for(int i = ; i < n; i++) /*每个骨头的重量*/
scanf("%d",&nm[i]); for(int i = ; i < n; i++)
for(int j = v; j >= nm[i];j--) /*dp算法*/
dp[j] = max(dp[j],dp[j-nm[i]]+nv[i]);
cout<<dp[v]<<endl;
}
} return ;
}
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