HDU 2602 Bone Collector(01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51035 Accepted Submission(s): 21446

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

01背包：

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>#include<math.h>using namespace std;int dp[1100];struct node{    int value,volume;} a[1100];int main(){    int t;    cin>>t;    while(t--)    {        int n,v,i,j;        memset(dp,0,sizeof(dp));        cin>>n>>v;        for(i=1; i<=n; i++)        {            cin>>a[i].value;        }        for(i=1;i<=n;i++)        {            cin>>a[i].volume;        }        for(i=1; i<=n; i++)        {            for(j=v; j>=a[i].volume; j--)            {                dp[j]=max(dp[j],dp[j-a[i].volume]+a[i].value);            }        }        cout<<dp[v]<<endl;    }    return 0;}