Hdu 2602 Bone Collector(简单01背包)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 54964    Accepted Submission(s): 22985

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

 

Author

Teddy

 

Source

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

 

题目分析：简单的01背包，用DP【i】表示当前背包里面还剩下的容量，用一维数组解可能好一点。不懂得请参考背包九讲。

代码如下：

#include<iostream>#include<cstring>#include<algorithm>const int maxx=1000+50;using namespace std;int dp[maxx],val[maxx],vo[maxx];int main(){int t;cin>>t;int N,V;while(t--){    cin>>N>>V;    memset(dp,0,sizeof(dp));    for(int i=0;i<N;i++)cin>>val[i];    for(int j=0;j<N;j++)cin>>vo[j];    for(int i=0;i<N;i++)    {        for(int j=V;j>=vo[i];j--)        {            dp[j]=max(dp[j],dp[j-vo[i]]+val[i]);        }    }cout<<dp[V]<<endl;}return 0;}