HDU2602--Bone Collector--动态规划

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本文探讨了一道经典的背包问题变种——骨收集者最大价值问题。在一个限定容量的背包中，如何选择不同体积和价值的骨头以达到价值最大化。文章提供了一个详细的编程解决方案，并附带了完整的代码实现。

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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

#include <iostream>
#include <cstdio>
using namespace std;
int dp[1008][1008];
int value[1008];
int vol[1008];
int max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,v;
		memset(dp,0,sizeof(dp));
		scanf("%d%d",&n,&v);
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&value[i]);
		}
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&vol[i]);
		}
		for(int i=1;i<=n;i++)//石头个数
		{
			for(int j=0;j<=v;j++)//背包体积不断变大
			{
				dp[i][j]=dp[i-1][j];
				if(j>=vol[i])
				{
					dp[i][j]=max(dp[i][j],dp[i-1][j-vol[i]]+value[i]);
				}
			}
		}
		cout<<dp[n][v]<<endl;
	}
	return 0;
}