hdu 2602 Bone Collector (01背包)
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
//最简单的01背包问题
#include <iostream> #include <cstring> #include <algorithm> using namespace std; int main() { int t,n,v; int value[1005],volume[1005],dp[1005]; cin>>t; while(t--) { cin>>n>>v; for(int i=1;i<=n;i++) cin>>value[i]; for(int j=1;j<=n;j++) cin>>volume[j]; memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) for(int j=v;j>=volume[i];j--) dp[j]=max(dp[j],dp[j-volume[i]]+value[i]); cout<<dp[v]<<endl; } return 0; }
你若是天才,我便是疯子