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发布时间:2022-07-27 13:55:35
 

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 21417    Accepted Submission(s): 8613

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 1 5 10 1 2 3 4 5 5 4 3 2 1   Sample Output 14  
#include<iostream>
#include<cstring>
#include<string>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,m,i,j,value[1020],volume[1020],dp[1020];
		scanf("%d %d",&n,&m);
		for(i=0;i<n;i++)
		{
			scanf("%d",&value[i]);
		}
		for(i=0;i<n;i++)
		{
			scanf("%d",&volume[i]);
		}
		memset(dp,0,sizeof(dp));
		for(i=0;i<n;i++)
		{
			for(j=m;j>=volume[i];j--)
			{
				dp[j] = dp[j-volume[i]] + value[i] > dp[j] ? dp[j-volume[i]]+value[i] : dp[j];
				//if((dp[j-v]))
			}
		}
		printf("%d
",dp[m]);
	}
	return 0;
}

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