B - Bone Collector HDU - 2602

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 ?

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Input

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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.

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Output

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One integer per line representing the maximum of the total value (this number will be less than 2 31).

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Sample Input

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

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Sample Output

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14

题解：了解了相关知识后，知道了这是个比较简单的01背包问题的裸题。我看了大佬的代码后，用二维数组始终没有实现，最终采用的是一维数组，发现的确比二维数组要简单，AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>                             //这个头文件一定不能少
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
	    int n,v;
	    int a[1005],b[1005],m[1005];
	    scanf("%d%d",&n,&v);
		memset(m,0,sizeof(m));                               //使数组m元素全为零
		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<n;i++)
		{
			for(int j=v;j>=b[i];j--)
			{
				m[j]=max(m[j],m[j-b[i]]+a[i]);                                 //选择最大的价值比
			}
		}
		printf("%d\n",m[v]);
	}
	return 0;
}

小白心得