HDU2602 Bone Collector【01背包】

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Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 86567    Accepted Submission(s): 35776

 

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

1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

14

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

问题链接：HDU2602 Bone Collector

解题思路：01背包模板题

AC的C++代码：

#include<iostream>
#include<algorithm>
#include<cstring>using namespace std;const int N=1005;
int c[N],p[N];//分别存储费用和价值
int dp[N]; int main()
{int t,N,V;scanf("%d",&t);while(t--){scanf("%d%d",&N,&V);for(int i=0;i<N;i++)scanf("%d",&p[i]);for(int i=0;i<N;i++)scanf("%d",&c[i]);memset(dp,0,sizeof(dp));for(int i=0;i<N;i++)for(int j=V;j>=c[i];j--)dp[j]=max(dp[j],dp[j-c[i]]+p[i]);printf("%d\n",dp[V]);}return 0;} 

 

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