Hdu2602 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
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int t,n,W; 5 ll v[1005]; 6 int w[1005]; 7 ll dp[1005][1005]; 8 int main() 9 { 10 cin>>t; 11 while(t--){ 12 cin>>n>>W; 13 for(int i=0;i<n;i++) cin>>v[i]; 14 for(int i=0;i<n;i++) cin>>w[i]; 15 for(int i=0;i<n;i++){ 16 for(int j=0;j<=W;j++){ 17 if(j<w[i]) dp[i+1][j]=dp[i][j]; 18 else dp[i+1][j]=max(dp[i][j],dp[i][j-w[i]]+v[i]); 19 } 20 } 21 cout<<dp[n][W]<<endl; 22 } 23 return 0; 24 }