hdu 2602（经典01背包）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 57334    Accepted Submission(s): 23933

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

01背包 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define N 10000
typedef long long ll;
using namespace std;
int value[N],vloume[N];
int dp[N];
int main()
{
    int t;
    scanf("%d",&t);
    int n,v;
    while(t--)
    {
        memset(value,0,sizeof(value));
        memset(vloume,0,sizeof(vloume));
        memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&v);
        for(int i=0;i<n;i++)
            scanf("%d",&value[i]);
        for(int i=0;i<n;i++)
            scanf("%d",&vloume[i]);
        for(int i=0;i<n;i++)
        {
            for(int j=v;j>=vloume[i];j--)
            {
                dp[j]=max(dp[j],dp[j-vloume[i]]+value[i]);
            }
        }
        printf("%d\n",dp[v]);
    }
}