[ACM] hdu 2602 Bone Collector(01背包)
Bone Collector
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25042 Accepted Submission(s): 10147
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
解题思路:
纯粹的01背包。
先附上二维数组的解法代码:
#include <iostream> #include <string.h> #include <algorithm> using namespace std; int dp[1002][1002]; int value[1002],w[1002]; int N,V; int main() { int t;cin>>t; while(t--) { cin>>N>>V; for(int i=1;i<=N;i++) cin>>value[i]; for(int i=1;i<=N;i++) cin>>w[i]; memset(dp,0,sizeof(dp)); for(int i=1;i<=N;i++) for(int j=0;j<=V;j++)//千万要记得重量要从0开始,不是1 { if(w[i]<=j) dp[i][j]=max(dp[i-1][j],dp[i-1][j-w[i]]+value[i]); else dp[i][j]=dp[i-1][j]; } cout<<dp[N][V]<<endl; } return 0; }
用一维数组写的代码:
#include <iostream> #include <string.h> #include <algorithm> using namespace std; int dp[1002]; int v[1002];//价值 int w[1002];//容量 int N,V;//骨头个数,最大容量 void solve()//01背包实现 { memset(dp,0,sizeof(dp)); for(int i=0;i<N;i++) for(int j=V;j>=w[i];j--) dp[j]=max(dp[j],dp[j-w[i]]+v[i]); } int main() { int t;cin>>t; while(t--) { cin>>N>>V; for(int i=0;i<N;i++) cin>>v[i]; for(int i=0;i<N;i++) cin>>w[i]; solve(); cout<<dp[V]<<endl; } return 0; }
运行: