hdu2602-01背包

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 48440 Accepted Submission(s): 20204

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
简单的01背包

#include <iostream>#include <cstdio>#include <cstring>using namespace std; int vol[1010],val[1010],dp[1010];int main(){    int t,n,v,i,j;    //freopen("in.txt","r",stdin);    while(scanf("%d",&t)!=EOF){        while(t--){            memset(dp,0,sizeof(dp));            scanf("%d%d",&n,&v);            for ( i = 0; i < n; ++i)            {                scanf("%d",&val[i]);            }            for ( i = 0; i < n; ++i)            {                scanf("%d",&vol[i]);            }               for ( i = 0; i < n; ++i)                    {                        for ( j = v; j >=vol[i]; j--)                        {                            dp[j]=max(dp[j],dp[j-vol[i]]+val[i]);                        }                    }               printf("%d\n", dp[v]);          }    }    return 0;}