HDU 2602 Bone Collector

最新推荐文章于 2020-04-02 16:55:44 发布

下饭厨子

最新推荐文章于 2020-04-02 16:55:44 发布

阅读量276

点赞数

CC 4.0 BY-SA版权

分类专栏： DP 背包

本文链接：https://blog.youkuaiyun.com/FlytobuDu/article/details/52209103

DP 同时被 2 个专栏收录

3 篇文章

订阅专栏

背包

1 篇文章

订阅专栏

本文介绍了一个经典的01背包问题实例——骨收集者问题，并详细解释了如何通过动态规划求解最大价值。输入包括测试案例数量、每种骨头的价值与体积，输出为骨收集者能获得的最大总价值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Bone Collector

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 51788 Accepted Submission(s): 21809

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

Sample Output

//简单 01背包主要便于理解01背包的思想；

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int t,N,V;
	int c[1010],w[1010];
	int dp[1010];
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&N,&V);
		for(int i=1;i<=N;i++)
		scanf("%d",&w[i]);
		for(int i=1;i<=N;i++)
		scanf("%d",&c[i]);
		memset(dp,0,sizeof(dp));
		for(int i=1;i<=N;i++)
		{
			for(int j=V;j>=c[i];j--)
			{
				dp[j]=max(dp[j],dp[j-c[i]]+w[i]);
			}
		}
		printf("%d\n",dp[V]);
	}
}