背包——Bone Collect

最新推荐文章于 2024-02-26 21:34:40 发布

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#背包

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本文探讨了在一个虚构小镇上，一个收集各种骨头的人如何利用有限的背包容量，通过优化策略来最大化收集到的骨头总价值的问题。通过给定的输入参数，包括骨头的数量、背包容量、每种骨头的价值和体积，读者可以学习到一种动态规划的方法来解决此类问题。

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 2 ³¹).
 
Sample Input
 1
5 10
1 2 3 4 5
5 4 3 2 1

Sample Output

#include"iostream"
#include"cstdio"
#include"cstring"
#include"cstdlib"
using namespace std;
int n,v;
int val[1050],V[1050],dp[1050][1050];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {memset(val,0,sizeof(val));
    memset(V,0,sizeof(V));
    memset(dp,0,sizeof(dp));
        scanf("%d%d",&n,&v);
        int i,j;
        for(i=1;i<=n;i++)
            scanf("%d",&val[i]);
             for(i=1;i<=n;i++)
                scanf("%d",&V[i]);
                for(int i=1;i<=n;i++)
                    for(j=0;j<=v;j++)
                {dp[i][j]=dp[i-1][j];
                   if(j>=V[i]) dp[i][j]=max(dp[i-1][j],dp[i-1][j-V[i]]+val[i]);
                }
                printf("%d\n",dp[n][v]);


    }
}