hdu 01背包问题 Bone Collector

最新推荐文章于 2021-03-24 11:19:10 发布

hpugym

最新推荐文章于 2021-03-24 11:19:10 发布

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分类专栏：背包问题 hdu oj

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本文探讨了一种经典的背包问题——骨收集者问题。通过分析不同骨头的价值与体积，利用背包算法实现价值最大化。介绍了使用贪心算法的尝试及其失败原因，并给出了正确的背包算法实现。

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Bone Collector

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 28 Accepted Submission(s) : 19

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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

Author

Teddy

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

解题报告：

题目大意就是说，收集骨头，每一种骨头有起固定的体积及价值，最终就是在固定体积的背包里怎么样能够获得最大的价值，开始时候想到用贪心（按照价值来排序，依次相取），但是一直WA，后来用起了背包，发现简单多了。但是解题的关键还是在于背包的使用。。。

//#include<stdio.h>
//#include<algorithm>
//#include<string.h>
//using namespace std;
//struct inin
//{
//	int val;
//	int vol;
//}boy[10010];
//int cmp(inin a,inin b)
//{
//	return a.val>b.val;
//}
//int main()
//{
//	int T;
//	int sum;
//	int n,Vol;
//	int i,j;
//	scanf("%d",&T);
//	while(T--)
//	{
//		scanf("%d%d",&n,&Vol);
//		for(i=0;i<n;i++)
//		  scanf("%d",&boy[i].val);
//		for(j=0;j<n;j++)
//		  scanf("%d",&boy[j].vol);
//		sort(boy,boy+n,cmp);
//		sum=0;
//		for(i=0;i<n;i++)
//		{
//			if(Vol-boy[i].vol>=0)
//			{
//				sum+=boy[i].val;
//				Vol-=boy[i].vol;
//			}
//			else
//			{
//				break;
//			}
//		} 
//		printf("%d\n",sum);
//	} 
//return 0;
//} 
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct inin
{
	int val;
	int vol;
}boy[10010];
int main()
{
	int T;
	int sum;
	int n,Vol;
	int i,j;
	int bag[10010];
	scanf("%d",&T);
	while(T--)
	{
		memset(boy,0,sizeof(boy));
		memset(bag,0,sizeof(bag));
		scanf("%d%d",&n,&Vol);
		for(i=0;i<n;i++)
		  scanf("%d",&boy[i].val);
		for(j=0;j<n;j++)
		  scanf("%d",&boy[j].vol);
		for(i=0;i<n;i++)
		{
			for(j=Vol;j>=boy[i].vol;j--)
			{
				<strong><span style="color:#ff0000;">bag[j]=max(bag[j],bag[j-boy[i].vol]+boy[i].val);</span></strong>
			}
		}
		printf("%d\n",bag[Vol]);
	} 
return 0;
}