HDOJ 2602 Bone Collector（背包问题）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 41716    Accepted Submission(s): 17357

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

//简单的背包问题

#include<cstring>#include<iostream>#define max(a,b) a>b?a:busing namespace std;struct node{int num,val;};node str[100100];int bag[100100];int main(){int m,n,i,j,t;cin>>t; while (t--){cin>>m>>n;memset(str,0,sizeof(str));memset(bag,0,sizeof(bag));for (i=0;i<m;i++)cin>>str[i].val;for (i=0;i<m;i++)cin>>str[i].num;for (i=0;i<m;i++){for (j=n;j>=str[i].num;j--){bag[j]=max(bag[j],bag[j-str[i].num]+str[i].val);}}cout<<bag[n];cout<<"\n";}return 0;}