Bone Collector
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 ?
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.
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
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 using namespace std; 5 6 int main() 7 { 8 int c[1005]; 9 int w[1005]; 10 int dp[1005]; 11 int t,V,v,i,n; 12 scanf("%d",&t); 13 while(t--) 14 { 15 memset(dp,0,sizeof(dp)); 16 scanf("%d%d",&n,&V); 17 for(i=1;i<=n;i++) 18 scanf("%d",&w[i]); 19 for(i=1;i<=n;i++) 20 scanf("%d",&c[i]); 21 for(i=1;i<=n;i++) 22 for(v=V;v>=0;v--) 23 { 24 if(v-c[i]>=0) 25 dp[v]=max(dp[v],dp[v-c[i]]+w[i]); 26 } 27 printf("%d ",dp[V]); 28 } 29 return 0; 30 }