杭电 oj2602~(0-1背包问题)
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51417 Accepted Submission(s): 21634
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
Author
Teddy
Source
简单的0-1背包问题 ~
#include<iostream> using namespace std; int main() { int t; int w[1050]; int v[1050]; int d[1050][1050]; while(cin>>t) { while(t--) { int n,c; cin>>n>>c; for(int i=1;i<=n;i++) cin>>w[i]; for(int i=1;i<=n;i++) cin>>v[i]; for(int i=1;i<=n;i++) for(int j=0;j<=c;j++) { d[i][j]=(i==1?0:d[i-1][j]); if(j>=v[i]) d[i][j]=max(d[i][j],d[i-1][j-v[i]]+w[i]); } cout<<d[n][c]<<endl; } } return 0; }