Bone Collector

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

1
5 10
1 2 3 4 5
5 4 3 2 1

示例输出

14

#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]);
        }
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值