Bone Collector

最新推荐文章于 2024-07-19 15:39:37 发布

lucky_vikey

最新推荐文章于 2024-07-19 15:39:37 发布

阅读量447

点赞数

CC 4.0 BY-SA版权

分类专栏： 01背包

本文链接：https://blog.youkuaiyun.com/lucky_vikey/article/details/38415337

01背包专栏收录该内容

4 篇文章

订阅专栏

探讨一个关于骨收集者如何在有限容量的背包中，通过收集不同价值和体积的骨头，最大化总价值的数学问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Bone Collector
题目描述

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 ?

输入

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.

输出

One integer per line representing the maximum of the total value (this number will be less than 231).

示例输入

示例输出

#include <iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int max(int n,int m)
{
    return n>m?n:m;
}
int main()
{
    int t,n,m,jia[1001],v[1001],f[1001],i,j,k;
    scanf("%d",&t);
    while(t--)
        {
            memset(f,0,sizeof(f));
            scanf("%d%d",&n,&m);
            for(i=0; i<n; i++)
                scanf("%d",&jia[i]);
            for(j=0; j<n; j++)
                scanf("%d",&v[j]);
            for(i=0; i<n; i++)//01背包
                for(j=m; j>=v[i]; j--)
                {
                    f[j]=max(f[j],f[j-v[i]]+jia[i]);
                }

        printf("%d\n",f[m]);
        }
}