Bone Collector（HDU-2602）

最新推荐文章于 2024-07-19 15:39:37 发布

原创最新推荐文章于 2024-07-19 15:39:37 发布 · 314 阅读

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#acm #背包

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本文介绍了一道经典的01背包问题，通过一个骨收集者的故事背景来阐述如何利用动态规划算法求解最大价值的问题。文章提供了完整的代码实现，并详细解释了输入输出样例。

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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

最简单的01背包

#include<iostream>
#include<cstring.>
#include<cstdio>
using namespace std;
int a[1005],b[1005],dp[1005];
int t,n,v;
int main()
{
	scanf("%d",&t);
	while(t--){
		scanf("%d%d",&n,&v);
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		for(int i=0;i<n;i++){
			scanf("%d",&b[i]);
		}
		for(int i=0;i<=v;i++){
			dp[i]=0;
		}
		for(int i=0;i<n;i++){
			for(int j=v;j>=b[i];j--){
				dp[j]=max(dp[j],dp[j-b[i]]+a[i]);
			}
		}
		printf("%d\n",dp[v]);
	}
    return 0;
}