Bone Collector（HDOJ2602)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20880    Accepted Submission(s): 8345

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
就现在的理解而言，我认为动态规划正如其名一样，就是不断以某种特定的条件书安心数组的值，此题是标准的背包问题，所以直接使用背包问题的公式即可得到所求的答案
代码：
#include<iostream>using namespace std;int c[1001],v[1001];int f[1010];int max(int a,int b){return a>b? a:b;}int main(){int T,i,j;cin>>T;int n,vo;while(T--){cin>>n>>vo;for(i=0;i<=vo;i++){f[i]=0;}for(i=1;i<=n;i++){cin>>v[i];}for(j=1;j<=n;j++){cin>>c[j];}for(i=1;i<=n;i++){for(j=vo;j>=c[i];j--){f[j]=max(f[j],f[j-c[i]]+v[i]);}}cout<<f[vo]<<endl;}return 0;}