杭电ACM 2014暑期集训队——选拔安排~



Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25800    Accepted Submission(s): 10453


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 ?
背包问题_i++



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<iostream>
using namespace std;
int max(int a,int b)
{if(a>b)
return a;
else return b;}
int n,j,v,c[1004],w[1004],dp[1004][1004],k,i;
int main()
{

cin>>k;
while(k--)
{
memset(dp,0,sizeof(dp));
cin>>n>>v;
for(i=1;i<=n;i++)
cin>>c[i];
for(i=1;i<=n;i++)
cin>>w[i];

for(i=1;i<=n;i++)
for(j=0;j<=v;j++)
if(w[i]<=j)
dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+c[i]);
else
dp[i][j]=dp[i-1][j];
cout<<dp[n][v]<<endl;
}
return 0;
}