HDOJ--2602--Bone Collector【背包问题】

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

 

Recommend

lcy

 

思路：01背包问题

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;int main(){    int n,v,t,i,j,value[1005],volume[1005],dp[1005];    scanf("%d",&t);    while(t--)    {        memset(value,0,sizeof(value));        memset(volume,0,sizeof(volume));        memset(dp,0,sizeof(dp));        scanf("%d%d",&n,&v);        for(i=1;i<=n;i++)            scanf("%d",&value[i]);        for(i=1;i<=n;i++)            scanf("%d",&volume[i]);        for(i=1;i<=n;i++)            for(j=v;j>=volume[i];j--)                if(dp[j]<dp[j-volume[i]]+value[i])                    dp[j]=dp[j-volume[i]]+value[i];        printf("%d\n",dp[v]);    }    return 0;}