HDU 2602 Bone Collector
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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
15 101 2 3 4 55 4 3 2 1
Sample Output
14#include <iostream>#include <cstring>using namespace std;int value[1010];int volume[1010];int dp[1010];int main(){ int T; int N,V; cin>>T; while(T--) { cin>>N>>V; for(int i=0; i<N; i++) cin>>value[i]; for(int i=0; i<N; i++) { cin>>volume[i]; } memset(dp,0,sizeof(dp));//清零; for(int i=0; i<N; i++) for(int j=V; j>=volume[i]; j--) { dp[j]=max(dp[j],dp[j-volume[i]]+value[i]);//思想核心; } cout<<dp[V]<<endl;//确保最后的体积是最大的时候输出; } return 0;}
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