hdu2602 Bone Collector(背包问题)

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 39612    Accepted Submission(s): 16412

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
   
   
   
   

 

Author

Teddy

 

Source

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

 

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Statistic |  Submit |  Discuss |  Note

非常常规的0 1背包问题。

第一排输入的是价值，第二排是体积

#include <stdio.h>
#include <string.h>
int main()
{
	int n,v;
	int ncase,dp[1005],val[1005],vol[1005];
	scanf("%d",&ncase);
	while(ncase--)
	{
		memset(dp,0,sizeof(dp));
		memset(val,0,sizeof(val));
		memset(vol,0,sizeof(vol));
		scanf("%d %d",&n,&v);
		for(int i=0;i<n;i++)
		scanf("%d",&val[i]);
		for(int i=0;i<n;i++)
		scanf("%d",&vol[i]);
		int max=0;
		for(int i=0;i<n;i++)
		{
			for(int j=v;j>=vol[i];j--)
			{
				if(dp[j]<dp[j-vol[i]]+val[i])
				dp[j]=dp[j-vol[i]]+val[i];
				if(dp[j]>max)
				max=dp[j];
			}
		}
		printf("%d\n",max);
	}
}