hdu2602 Bone Collector（01背包）

最新推荐文章于 2021-01-18 14:47:36 发布

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本文详细解析了一个经典的01背包问题案例，介绍了如何通过动态规划算法解决该问题，并提供了完整的代码实现。

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Bone Collector

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

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最最最最最最最经典的01背包问题，告诉你物品数量和背包容量及各件物品的质量和价值，求最大价值

我们来考虑第i件物品的决策情况

若容量够可以取，则考虑取了之后最大价值和不取的最大价值何者大，即dp[i][j]=max(dp[i][j-1],dp[i-weight[j]][j-1]+val[j]);
若容量不够不能取，则dp[i][j]=dp[i][j-1];

代码入下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int dp[1005][1005];
int main()
{
	int weight[1005],val[1005],sum,n,o;
	while(~scanf("%d",&o))
	{
		while(o--)
		{
			scanf(" %d %d",&n,&sum);
			for(int i=1;i<=n;i++)
				scanf("%d",&val[i]);
			for(int i=1;i<=n;i++)
				scanf("%d",&weight[i]);
			memset(dp,0,sizeof(dp));
			for(int i=0;i<=sum;i++)
				for(int j=1;j<=n;j++)
				{
					if(dp[i][j]+weight[j]<=i)
						dp[i][j]=max(dp[i][j-1],dp[i-weight[j]][j-1]+val[j]);
					else
						dp[i][j]=dp[i][j-1];			
				}
			printf("%d\n",dp[sum][n]);

		}
	}
	return 0;
}