hdu——2602Bone Collector（第一类背包问题）

Bone Collector

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 46744    Accepted Submission(s): 19463

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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第一类背包  for(i=1;i<=n;i++）

for(j=V;j>v[i];j--)

d[j]=max(d[j],d[j-v[i]]+jg[i])

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<set>
#include<map>
#include<sstream>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<deque>
using namespace std;
struct bb
{
	long long v;
	long long jz;
}a[1005];
long long max(long long n,long long m)
{
	if(n>=m)return n;
	else return m;
}
int main()
{
	long long n,m,vv,sum[1005];
	cin>>n;
	while(n--)
	{	
		cin>>m>>vv;
		memset(sum, 0, sizeof(sum));
		for(int i=0;i<m;i++)
		{
			cin>>a[i].jz;
		}
		for(int i=0;i<m;i++)
		{
			cin>>a[i].v;
		}
		for(int i=0;i<m;i++)
		{
			for(int j=vv;j>=a[i].v;j--)
			{
				sum[j]=max(sum[j],sum[j-a[i].v]+a[i].jz);	
			}
		}
		cout<<sum[vv]<<endl;//注意是vv而不是其他
	}	
	return 0;
}