[hdu]Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 37393    Accepted Submission(s): 15446

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 2 ³¹).

 

Sample Input

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

 

Sample Output

14

 

题意：一个手机骨头的人要带着很多很多的骨头上路，但是包包装不下太多东西，问包包最多能装价值多少的骨头。

题解：01类背包问题，注意状态转移方程

#include "stdio.h"
#include "string.h"
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		int f[1001]={0};
		int N,V,val[1001],vol[1001];
		scanf("%d %d",&N,&V);
		for(int i=1;i<=N;i++)scanf("%d",&val[i]);//value
		for(int i=1;i<=N;i++)scanf("%d",&vol[i]);//volume
		for(int i=1;i<=N;i++)
		{
			for(int j=V;j>=vol[i];j--)
			{
				f[j]=f[j]>f[j-vol[i]]+val[i]?f[j]:f[j-vol[i]]+val[i];//01型背包的 状态转移方程  
			}
		}
		printf("%d\n",f[V]);
	}
	return 0;
}