hdu 2602 None Collector [01背包问题]

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本文介绍了一道典型的01背包问题，并提供了详细的解题思路及C语言实现代码。通过实例讲解了如何求解给定体积背包能容纳的最大价值物品组合。

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Bone Collector

hdu 2602 题目链接

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 2 31).

Sample Input

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

Sample Output

题意概括：

先在给你V体积的背包，然后有n块石头，每块石头有对应的体积大小和价值大小，请求出在背包体积下所能装最大价值的石头。

解题思路：

01背包问题，关键是对F[v]=MAX{F[v],F[v-c[i]]+w[i]};的理解。01背包，其实0代表否定1代表肯定，由名字就知道是是与否的问题，也就是对这块石头是否放进去的考虑。v是现有体积，要想把第i块石头放进去，就必须留有c[i]个体积，所以之前最多装有v-c[i]个体积，然后就是比较之前的体积的最大价值和放入这块石头之后价值总和的最大价值，因为每一步都保证就是该体积的最大价值，所以就这样过一遍就好了。

程序代码：

#include <stdio.h>
#include <string.h>
int f[2100],v[1010],c[1010];
int max(int a,int b)
{
	return a>b?a:b;
}

int main()
{
	int t,m,n,i,j;
	scanf("%d",&t);
	while(t--)
	{
		memset(f,0,sizeof(f));
		memset(v,0,sizeof(v));
		memset(c,0,sizeof(c));
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
			scanf("%d",&c[i]);
		for(i=1;i<=n;i++)
			scanf("%d",&v[i]);
		for(i=1;i<=n;i++)
			for(j=m;j>=v[i];j--)
				f[j]=max(f[j],f[j-v[i]]+c[i]);
		printf("%d\n",f[m]);
	}
	return 0;
}