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.
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
题目大意:
给定n,v分别代表n个骨头,体积为v的背包,接下来n个数字代表n个骨头的价值,再接下来n个数字代表n个骨头的体积。
01背包模板
#include <iostream> #include <cstring> #include <algorithm> using namespace std; int w[1005],v[1005],dp[1005]; int n,m; int main() { int T; cin>>T; while(T--) { memset(dp,0,sizeof dp); cin>>n>>m; for(int i=1;i<=n;i++) cin>>v[i]; for(int i=1;i<=n;i++) cin>>w[i]; for(int i=1;i<=n;i++) for(int j=m;j>=w[i];j--) dp[j]=max(dp[j],dp[j-w[i]]+v[i]); cout<<dp[m]<<' '; } return 0; }