HDU 2602 Bone Collector

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 65814    Accepted Submission(s): 27430

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 ?

 

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.

 

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

 

题目大意：给你两个数字N和V，N是这个骨头的数量，V是背包的容量，第一行给出每个骨头的价值，第二行是每个骨头的体积。求背包能装的最大的价值。

01背包水题

#include<iostream>#include<cstring>using namespace std;int weight[1010],value[1010],dp[1010];int main(){int t;cin>>t;while(t--){memset(dp,0,sizeof(dp));int n,v;cin>>n>>v;for(int i=1;i<=n;i++) cin>>value[i];for(int i=1;i<=n;i++) cin>>weight[i];for(int i=1;i<=n;i++){for(int j=v;j>=weight[i];j--){dp[j]=max(dp[j],dp[j-weight[i]]+value[i]);}}cout<<dp[v]<<endl;}}