hduoj2602Bone Collector

简介:  Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 291...



Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29163    Accepted Submission(s): 11922


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
 翻译:有一人爱收集骨头,每个骨头有一个价值和重量,他有一个背包,容量为V,问他装得最大价值量是多少?(经典的01背包问题)

重点:01背包、动态规划
难点:动态方程的书写


#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
	int T,n,v,i,j,dp[1100],a[1100],b[1100];//a[1100]是价值,b[1100]是重量;
	cin>>T;
	while(T--)
	{
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b)); 
		memset(dp,0,sizeof(dp)); 
		cin>>n>>v;//n是骨头个数,V是背包容量 
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(i=1;i<=n;i++)
			scanf("%d",&b[i]);
		for(i=1;i<=n;i++)
			for(j=v;j>=b[i];j--)
			{
				if(dp[j]<dp[j-b[i]]+a[i])
					dp[j]=dp[j-b[i]]+a[i];
			}
		printf("%d\n",dp[v]);
	}
	return 0;
}

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