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 ?
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
模板题,找到状态转移方程,使用一维01背包,对于背包的体积要逆序遍历,这样是的右边的max(f[j],f[j-v[i]]+val[i])能够表示为上一层的状态,而对于当j<v[i时,默认f[j]不变,是上一层的值
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 using namespace std; 5 int main() 6 { 7 int T,V; 8 int v[1001],val[1001]; 9 int f[1001]; 10 int i,N,j; 11 cin>>T; 12 while(T--) 13 { 14 memset(f,0,sizeof(f)); 15 cin>>N>>V; 16 for(i=1;i<=N;i++) 17 cin>>val[i]; 18 for(i=1;i<=N;i++) 19 cin>>v[i]; 20 for(i=1;i<=N;i++) 21 { 22 for(j=V;j>=v[i];j--) 23 f[j]=max(f[j],f[j-v[i]]+val[i]); 24 } 25 cout<<f[V]<<endl; 26 } 27 }