HDU 2602 Bone Collector (简单的0-1背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24860    Accepted Submission(s): 10062

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

 

#include <cstdio>#include <cstdlib>#include <cstring>#include <climits>const int MAX = 1011;int value[MAX],cost[MAX],pack[MAX];void init(){int i;for(i=0;i<MAX;++i)pack[i] = 0;}void max(int &a,int b){if(a<b) a = b;}void zeroOnePack(int V,int n){int i,v;for(i=0;i<n;++i){for(v=V;v>=cost[i];--v){max(pack[v],pack[v-cost[i]]+value[i]);}}}int main(){int t,n,all,i;//freopen("in.txt","r",stdin);scanf("%d",&t);while(t--){init();scanf("%d %d",&n,&all);for(i=0;i<n;++i){scanf("%d",&value[i]);}for(i=0;i<n;++i){scanf("%d",&cost[i]);}zeroOnePack(all,n);printf("%d\n",pack[all]);}return 0;}